6  Sequences in \(\mathbb{R}\)

A sequence is an infinite list of real numbers. For example, the following are sequences:

• \((1,2,3,4, \ldots)\)

• \((-1,1,-1,1, \ldots)\)

• \(\left(1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \ldots\right)\)

Remark 1

• The order of elements in a sequence matters.

For example \[ (1,2,3,4,5,6, \ldots) \neq(2,1,4,3,6,5, \ldots) \]

• A sequence is not a set.

For example \[ \{-1,1,-1,1,-1,1, \ldots\}=\{-1,1\} \] but we cannot make a similar statement for the sequence \[ (-1,1,-1,1,-1,1, \ldots) \,. \]

• The above notation is ambiguous.

For example the sequence \[ (1,2,3,4, \ldots) \] can continue as

\[ (1,2,3,4,1,2,3,4,1,2,3,4, \ldots) \,. \]

• In the sequence \[ \left(1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \ldots\right) \] the elements get smaller and smaller, and closer and closer to \(0\). We say that this sequence converges to \(0\), or has \(0\) as a limit.

We would like to make the notions of sequence and convergence more precise.

6.1 Definition of sequence

We start with the definition of sequence of Real numbers.

Definition 2: Sequence of Real numbers

A sequence \(a\) in \(\mathbb{R}\) is a function \[ a \colon \mathbb{N}\to \mathbb{R}\,. \] For \(n \in \mathbb{N}\), we denote the \(n\)-th element of the sequence \(a\) by \[ a_{n}=a(n) \] and write the sequence as \[ \left(a_{n}\right)_{n \in \mathbb{N}} \,. \]

Notation 3

We will sometimes omit the subscript \(n \in \mathbb{N}\) and simply write \[ \left(a_{n}\right) \,. \] In certain situations, we will also write \[ \left(a_{n}\right)_{n=1}^{\infty} \,. \]

Example 4

• In general \(\left(a_{n}\right)_{n \in \mathbb{N}}\) is the sequence \[ \left(a_{1}, a_{2}, a_{3}, \ldots\right)\,. \]

• Consider the function \[ a \colon \mathbb{N}\to \mathbb{N}\, , \quad n \mapsto 2 n \,. \] This is also a sequence of real numbers. It can be written as \[ (2 n)_{n \in \mathbb{N}} \] and it represents the sequence of even numbers \[ (2,4,6,8,10, \ldots) \,. \]

• Let \[ a_{n}=(-1)^{n} \] Then \(\left(a_{n}\right)\) is the sequence \[ (-1,1,-1,1,-1,1, \ldots) \,. \]

• \(\left(\frac{1}{n}\right)_{n \in \mathbb{N}}\) is the sequence \[ \left(1, \frac{1}{2}, \frac{1}{4}, \frac{1}{5}, \ldots\right) \,. \]

6.2 Convergent sequences

We have notice that the sequence \(\left(\frac{1}{n}\right)_{n \in \mathbb{N}}\) gets close to \(0\) as \(n\) gets large. We would like to say that \(a_{n}\) converges to \(0\) as \(n\) tends to infinity.

To make this precise, we first have to say what it means for two numbers to be close. For this we use the notion of absolute value, and say that:

• \(x\) and \(y\) are close if \(|x-y|\) is small.
• \(|x-y|\) is called the distance between \(x\) and \(y\)
• For \(x\) to be close to \(0\), we need that \(|x-0|=|x|\) is small.

Saying that \(|x|\) is small is not very precise. Let us now give the formal definition of convergent sequence.

Definition 5: Convergent sequence

The real sequence \((a_n)\) converges to \(a\), or equivalently has limit \(a\), denoted by \[ \lim_{n \rightarrow \infty} a_{n}=a \,, \] if for all \(\varepsilon\in \mathbb{R}, \varepsilon>0\), there exists \(N \in \mathbb{N}\) such that for all \(n \in \mathbb{N}, n \geq N\) it holds that \[ \left|a_{n}-a\right| < \varepsilon\,. \] Using quantifiers, we can write this as \[ \forall \, \varepsilon>0 , \, \exists \, N \in \mathbb{N}\, \text{ s.t. } \, \forall \, n \geq N \,, \,\left|a_{n}-a\right| < \varepsilon\,. \] The sequence \(\left(a_{n}\right)_{n \in \mathbb{N}}\) is convergent if it admits limit.

Notation 6

We will often write \[ a_n \to a \] in place of \[ \lim_{n \rightarrow \infty} a_{n}=a \,. \]

Remark 7

• Informally, Definition 129 says that, no matter how small we choose \(\varepsilon\) (as long as it is strictly positive), we always have that \(a_{n}\) has a distance to \(a\) of less than or equal to \(\varepsilon\) from a certain point onwards (i.e., from \(N\) onward). The sequence \(\left(a_{n}\right)\) may fluctuate wildly in the beginning, but from \(N\) onward it should stay within a distance of \(\varepsilon\) of \(a\).

• In general \(N\) depends on \(\varepsilon\). If \(\varepsilon\) is chosen smaller, we might have to take \(N\) larger: this means we need to wait longer before the sequence stays within a distance \(\varepsilon\) from \(a\).

We now prove that the sequence \[ a_n = \frac1n \] converges to \(0\), according to Definition 129.

Theorem 8

The sequence \(\left(\frac{1}{n}\right)_{n \in \mathbb{N}}\) converges to \(0\) , i.e.,

\[ \lim_{n \rightarrow \infty} \frac{1}{n}=0 \,. \]

We give two proofs of the above theorem:

• Long proof, with all the details.
• Short proof, with less details, but still acceptable.

Proof: Proof of Theorem 8 (Long version)

We have to show that

\[ \lim _{n \rightarrow \infty} \frac{1}{n}=0, \]

which by definition is equivalent to showing that

\[ \forall \, \varepsilon>0 \,, \, \exists \, N \in \mathbb{N}\, \text{ s.t. } \, \forall \, n \geq N \,, \, \left|\frac{1}{n}-0\right| < \varepsilon\,. \tag{6.1}\]

Let \(\varepsilon\in \mathbb{R}\) with \(\varepsilon>0\). Choose \(N \in \mathbb{N}\) such that \[ N > \frac{1}{\varepsilon}\,. \] Such natural number \(N\) exists thanks to the Archimedean property. The above implies \[ \frac{1}{N} < \varepsilon\,, \tag{6.2}\] Let \(n \in \mathbb{N}\) with \(n \geq N\). By (6.2) we have \[ \frac{1}{n} \leq \frac{1}{N} < \varepsilon\,. \] From this we deduce \[ \left|\frac{1}{n}-0\right|=\frac{1}{n} \leq \frac{1}{N} < \varepsilon \]

Since \(n \in \mathbb{N}, n \geq N\) was arbitrary, we have proven that \[ \left|\frac{1}{n}-0\right| < \varepsilon\,, \quad \forall \, n \geq N \,. \tag{6.3}\] Condition (6.3) holds for all \(\varepsilon>0\), for the choice of \(N \in \mathbb{N}\) such that \[ N > \frac{1}{\varepsilon} \,. \] We have hence shown (6.1), and the proof is concluded.

As the above proof is quite long and includes lots of details, it is acceptable to shorten it. For example:

• We skip some intermediate steps.
• We do not mention the Archimedean property.
• We leave out the conclusion when it is obvious that the statement has been proven.

Proof: Proof of Theorem 8 (Short version)

We have to show that

\[ \forall \, \varepsilon>0 \,, \, \exists \, N \in \mathbb{N}\, \text{ s.t. } \, \forall \, n \geq N \,, \, \left|\frac{1}{n}-0\right| < \varepsilon\,. \]

Let \(\varepsilon>0\). Choose \(N \in \mathbb{N}\) such that \[ N > \frac{1}{\varepsilon}\,. \] Let \(n \geq N\). Then \[ \left|\frac{1}{n}-0\right|=\frac{1}{n} \leq \frac{1}{N} < \varepsilon\,. \]

In Theorem 8 we showed that \[ \lim_{n \to \infty} \, \frac{1}{n} = 0 \,. \] We can generalise this statement to prove that \[ \lim_{n \to \infty} \, \frac{1}{n^{p}} = 0 \] for any \(p>0\) fixed.

Theorem 9

For all \(p>0\), the sequence \(\left(\frac{1}{n^{p}}\right)_{n \in \mathbb{N}}\) converges to \(0\) , i.e.,

\[ \lim _{n \rightarrow \infty} \frac{1}{n^{p}}=0 \]

Proof

Let \(p>0\). We have to show that \[ \forall \varepsilon>0 \,, \, \exists \, N \in \mathbb{N}\, \text{ s.t. } \, \forall \, n \geq N \,, \, \left|\frac{1}{n^{p}}-0\right| < \varepsilon\,. \] Let \(\varepsilon>0\). Choose \(N \in \mathbb{N}\) such that \[ N > \frac{1}{\varepsilon^{1 / p}} \,. \tag{6.4}\] Let \(n \geq N\). Since \(p>0\), we have \(n^{p} \geq N^{p}\), which implies \[ \frac{1}{n^p} \leq \frac{1}{N^p} \,. \] By (A.29) we deduce \[ \frac{1}{N^p} < \varepsilon\,. \] Then \[ \left|\frac{1}{n^{p}}-0\right|=\frac{1}{n^{p}} \leq \frac{1}{N^{p}} < \varepsilon\,. \]

Question 10

Why did we choose \(N \in \mathbb{N}\) such that \[ N > \frac{1}{\varepsilon^p} \] in the above proof?

The answer is: because it works. Finding a number \(N\) that makes the proof work requires some rough work: Specifically, such rough work consists in finding \(N \in \mathbb{N}\) such that the inequality \[ |a_N - a | < \varepsilon \] is satisfied.

Important

Any rough work required to prove convergence must be shown before the formal proof (in assignments).

Example 11

Question. Using the definition of convergence, prove that \[ \lim_{n \to \infty} \frac{n}{2n+3} = \frac{1}{2} \,. \]

Solution.

1. Rough Work: Let \(\varepsilon>0\). We want to find \(N \in \mathbb{N}\) such that \[ \left|\frac{n}{2n+3}-\frac{1}{2}\right| < \varepsilon\,, \quad \forall \, n \geq N \,. \] To this end, we compute: \[\begin{align*} \left|\frac{n}{2 n+3}-\frac{1}{2}\right| & =\left|\frac{2n -(2n + 3)}{2(2n +3)}\right| \\ & =\left|\frac{- 3}{4n + 6}\right| \\ & = \frac{3}{4n + 6} \,. \end{align*}\] Therefore \[\begin{align*} \left|\frac{n}{2n+3}-\frac{1}{2}\right| < \varepsilon \quad \iff & \quad \frac{3}{4n + 6} < \varepsilon\\ \quad \iff & \quad \frac{4n + 6}{3} > \frac{1}{\varepsilon} \\ \quad \iff & \quad 4n + 6 > \frac{3}{\varepsilon} \\ \quad \iff & \quad 4n > \frac{3}{\varepsilon} - 6 \\ \quad \iff & \quad n > \frac{3}{4\varepsilon} - \frac{6}{4} \,. \end{align*}\] Looking at the above equivalences, it is clear that \(N \in \mathbb{N}\) has to be chosen so that \[ N > \frac{3}{4\varepsilon} - \frac{6}{4} \,. \]

2. Formal Proof: We have to show that \[ \forall \varepsilon>0 \,, \, \exists \, N \in \mathbb{N}\, \text{ s.t. } \, \forall \, n \geq N \,, \, \left|\frac{n}{2n+3}-\frac{1}{2}\right| < \varepsilon\,. \] Let \(\varepsilon>0\). Choose \(N \in \mathbb{N}\) such that \[ N > \frac{3}{4\varepsilon} - \frac{6}{4} \,. \tag{6.5}\] By the rough work shown above, inequality (A.30) is equivalent to \[ \frac{3}{4N + 6} < \varepsilon\,. \] Let \(n \geq N\). Then \[\begin{align*} \left|\frac{n}{2n+3}-\frac{1}{2}\right| & = \frac{3}{4 n+ 6 } \\ & \leq \frac{3}{4 N+ 6 } \\ & < \varepsilon\,, \end{align*}\] where in the third line we used that \(n \geq N\).

We conclude by showing that constant sequences always converge.

Theorem 12

Let \(c \in \mathbb{R}\) and define the constant sequence \[ a_n := c \,, \quad \forall \, n \in \mathbb{N}\,. \] We have that \[ \lim_{n \to \infty} \, a_n = c \,. \]

Proof

We have to prove that \[ \forall \varepsilon>0 \,, \, \exists \, N \in \mathbb{N}\, \text{ s.t. } \, \forall \, n \geq N \,, \, \left|a_n - c \right| < \varepsilon\,. \tag{6.6}\] Let \(\varepsilon>0\). We have \[ \left|a_n - c \right| = \left|c - c \right| = 0 < \varepsilon\,, \quad \forall \, n \in \mathbb{N}\,. \] Therefore we can choose \(N=1\) and (6.6) is satisfied.

6.3 Divergent sequences

The opposite of convergent sequences are divergent sequences.

Definition 13: Divergent sequence

We say that a sequence \(\left(a_{n}\right)_{n \in \mathbb{N}}\) in \(\mathbb{R}\) is divergent if it is not convergent.

Remark 14

Proving that a sequence \((a_n)\) is divergent is more complicated than showing it is convergent: To show that \((a_n)\) is divergent, we need to show that \((a_n)\) cannot converge to \(a\) for any \(a \in \mathbb{R}\).

In other words, we have to show that there does not exist an \(a \in \mathbb{R}\) such that \[ \lim _{n \rightarrow \infty} \, a_n = a \,. \] Using quantifiers, this means \[ \nexists \, a \in \mathbb{R}\, \text{ s.t. } \, \forall \, \varepsilon>0 \,, \, \exists \, N \in \mathbb{N}\, \text{ s.t. } \, \forall \, n \geq N \,, \, \left|a_{n}-a\right| < \varepsilon\,. \] The above is equivalent to showing that \[ \forall \, a \in \mathbb{R}\,, \, \exists \, \varepsilon> 0 \, \text{ s.t. } \, \forall \, N \in \mathbb{N}\,, \, \exists \, n \geq N \, \text{ s.t. } \, \left|a_{n}-a\right| \geq \varepsilon\,. \]

Theorem 15

Let \(\left(a_{n}\right)\) be the sequence defined by \[ a_{n}=(-1)^{n} \,. \] Then \(\left(a_{n}\right)\) does not converge.

Proof

To prove that \(\left(a_{n}\right)\) does not converge, we have to show that \[ \forall \, a \in \mathbb{R}\,, \, \exists \, \varepsilon> 0 \, \text{ s.t. } \, \forall \, N \in \mathbb{N}\,, \, \exists \, n \geq N \, \text{ s.t. } \, \left|a_{n}-a\right| \geq \varepsilon\,. \] Let \(a \in \mathbb{R}\). Choose \[ \varepsilon=\frac{1}{2} \,. \] Let \(N \in \mathbb{N}\). We distinguish two cases:

• \(a \geq 0\): Choose \(n=2 N+1\). Note that \(n \geq N\). Then \[\begin{align*} \left|a_{n}-a\right| & = \left|a_{2 N+1}-a\right| \\ & = \left|(-1)^{2 N+1}-a\right| \\ & =|-1-a| \\ & =1+a \\ & \geq 1 \\ & > \frac{1}{2} = \varepsilon\,, \end{align*}\] where we used that \(a \geq 0\), and therefore \[ |-1-a|=1+a \geq 1 \,. \]

• \(a<0\): Choose \(n=2 N\). Note that \(n \geq N\). Then \[\begin{align*} \left|a_{n}-a\right| & = \left|a_{2 N}-a\right| \\ & =\left|(-1)^{2 N}-a\right| \\ & = |1-a| \\ & = 1-a \\ & > 1 \\ & > \frac{1}{2} = \varepsilon\,, \end{align*}\] where we used that \(a < 0\), and therefore \[ |1-a|=1-a > 1 \,. \]

We propose an alternative proof by contradiction.

Proof: Alternative proof to Theorem 133

Suppose by contradiction that \(a_n \to a\) for some \(a \in \mathbb{R}\). Let \[ \varepsilon:=\frac12 \,. \] Since \(a_n \to a\), there exists \(N \in \mathbb{N}\) such that \[ |a_n - a| < \varepsilon= \frac13 \, \quad \forall \, n \geq N \,. \] If we take \(n = 2N\), then \(n \geq N\) and \[ |a_{2N} - a| = | 1 - a | < \frac12 \,. \] If we take \(n = 2N + 1\), then \(n \geq N\) and \[ |a_{2N+1} - a| = | -1 - a | < \frac12 \,. \] Therefore \[\begin{align*} 2 & = | (1 - a) - (-1 - a) | \\ & \leq |1 - a| + |-1 - a| \\ & < \frac12 + \frac12 = 1 \,, \end{align*}\] which is a contradiction. Hence \((a_n)\) is divergent.

6.4 Uniqueness of limit

In Definition 129 of convergence, we used the notation \[ \lim_{n \rightarrow \infty} a_{n}=a \,. \] The above notation makes sense only if the limit is unique, that is, if we do not have that
\[ \lim_{n \rightarrow \infty} a_{n} = b \,, \] for some \[ a \neq b \,. \] In the next theorem we will show that the limit is unique, if it exists.

Theorem 16: Uniqueness of limit

Let \(\left(a_{n}\right)_{n \in \mathbb{N}}\) be a sequence. Suppose that \[ \lim_{n \rightarrow \infty} a_{n} = a \,, \quad \lim_{n \rightarrow \infty} a_{n} = b \,. \] Then \(a=b\).

Proof

Assume that, \[ \lim_{n \rightarrow \infty} a_{n} = a \,, \quad \lim_{n \rightarrow \infty} a_{n} = b \,. \] Suppose by contradiction that \[ a \neq b \,. \] Choose \[ \varepsilon:= \frac{1}{2} \, |a-b| \,. \] Therefore \(\varepsilon>0\), since \(|a-b|>0\). By the convergence \(a_{n} \rightarrow a\), \[ \exists \, N_1 \in \mathbb{N}\, \text{ s.t. } \, \forall \, n \geq N_1 \,, \, \left|a_{n}-a\right| < \varepsilon\,. \] By the convergence \(a_{n} \rightarrow b\), \[ \exists \, N_2 \in \mathbb{N}\, \text{ s.t. } \, \forall \, n \geq N_2 \,, \, \left|a_{n} - b \right| < \varepsilon\,. \] Define \[ N := \max \{ N_1, N_2 \} \,. \] Choose an \(n \in \mathbb{N}\) such that \(n \geq N\). In particular \[ n \geq N_1 \,, \quad n \geq N_2 \,. \] Then \[\begin{align*} 2 \varepsilon& = |a-b| \\ & = \left|a-a_{n}+a_{n}-b\right| \\ & \leq\left|a-a_{n}\right|+\left|a_{n}-b\right| \\ & < \varepsilon+ \varepsilon\\ & = 2 \varepsilon\,, \end{align*}\] where we used the triangle inequality in the first inequality. Hence \(2 \varepsilon< 2 \varepsilon\), which gives a contradiction.

Example 17

Question. Prove that \[ \lim _{n \rightarrow \infty} \frac{n^{2}-1}{2 n^{2}-3}=\frac{1}{2} \]

Solution. According to Theorem 134, it suffices to show that the sequence \[ \left(\frac{n^{2}-1}{2 n^{2}-3}\right)_{n \in \mathbb{N}} \] converges to \(\frac{1}{2}\), since then \(\frac{1}{2}\) can be the only limit.

1. Rough Work: Let \(\varepsilon>0\). We want to find \(N \in \mathbb{N}\) such that \[ \left|\frac{n^{2}-1}{2 n^{2}-3} - \frac{1}{2}\right| < \varepsilon\,, \quad \forall \, n \geq N \,. \] To this end, we compute: \[\begin{align*} \left|\frac{n^{2}-1}{2 n^{2}-3}-\frac{1}{2}\right| & =\left|\frac{2\left(n^{2}-1\right)-\left(2 n^{2}-3\right)}{2\left(2 n^{2}-3\right)}\right| \\ & =\left|\frac{1}{4 n^{2}-6}\right| \\ & =\frac{1}{4 n^{2}-6} \\ & =\frac{1}{3 n^{2}+n^{2}-6} \\ & \leq \frac{1}{3 n^{2}} \end{align*}\] which holds if \(n \geq 3\), since in this case \(n^2 - 6 \geq 0\). Therefore \[\begin{align*} \left|\frac{n^{2}-1}{2 n^{2}-3}-\frac{1}{2}\right| < \varepsilon \quad \impliedby & \quad \frac{1}{3n^2} < \varepsilon\\ \quad \iff & \quad 3n^2 > \frac{1}{\varepsilon} \\ \quad \iff & \quad n^2 > \frac{1}{3\varepsilon} \\ \quad \iff & \quad n > \frac{1}{\sqrt{3\varepsilon}} \,. \end{align*}\] Looking at the above implications, it is clear that \(N \in \mathbb{N}\) has to be chosen so that \[ N > \frac{1}{\sqrt{3\varepsilon}} \,. \] Moreover we need to recall that \(N\) has to satisfy \[ N \geq 3 \] for the estimates to hold.

2. Formal Proof: We have to show that \[ \forall \varepsilon>0 \,, \, \exists \, N \in \mathbb{N}\, \text{ s.t. } \, \forall \, n \geq N \,, \, \left|\frac{n^{2}-1}{2 n^{2}-3}-\frac{1}{2}\right| < \varepsilon\,. \] Let \(\varepsilon>0\). Choose \(N \in \mathbb{N}\) such that \[ N > \max \left\{ \frac{1}{\sqrt{3\varepsilon}}, 3 \right\} \,. \] Let \(n \geq N\). Then \[\begin{align*} \left|\frac{n^{2}-1}{2 n^{2}-3}-\frac{1}{2}\right| & =\frac{1}{4 n^{2}-6} \\ & =\frac{1}{3 n^{2}+n^{2}-6} \\ & \leq \frac{1}{3 n^{2}} \\ & \leq \frac{1}{3 N^{2}} \\ & < \varepsilon\,, \end{align*}\] where we used that \[ n \geq N \geq 3 \] which implies \[ n^2 - 6 \geq 0\,, \] in the third line. The last inequality holds, since it is equivalent to \[ N > \frac{1}{\sqrt{3 \varepsilon}} \,. \]

6.5 Bounded sequences

An important property of sequences is boundedness.

Definition 18: Bounded sequence

A sequence \(\left(a_{n}\right)_{n \in \mathbb{N}}\) is called bounded if there exists a constant \(M \in \mathbb{R}\), with \(M>0\), such that \[ \left|a_{n}\right| \leq M \,, \quad \forall \, n \in \mathbb{N}\,. \]

Definition 135 says that a sequence is bounded, if we can find some constant \(M>0\) (possibly very large), such that for all elements of the sequence it holds that \[ \left|a_{n}\right| \leq M \,, \] or equivalently, that \[ -M \leq a_{n} \leq M \,. \]

We now show that any sequence that converges is also bounded

Theorem 19

Every convergent sequence is bounded.

Proof

Suppose the sequence \(\left(a_{n}\right)_{n \in \mathbb{N}}\) converges and let \[ a: = \lim_{n \rightarrow \infty} a_{n} \] By definition of convergence we have that \[ \forall \, \varepsilon>0 \,, \, \exists N \in \mathbb{N}\, \text{ s.t. } \, \forall \, n \geq N \,, \, \left|a_{n}-a \right| < \varepsilon\,. \] In particular, we can choose \[ \varepsilon= 1 \] and let \(N \in \mathbb{N}\) be that value such that \[ \left|a_{n}-a \right| < 1 \,, \quad \forall \, n \geq N \,. \] If \(n \geq N\) we have, by the triangle inequality, \[\begin{align*} \left|a_{n}\right| & = \left|a_{n}-a+a\right| \\ & \leq \left|a_{n}-a\right|+|a| \\ & < 1+|a| \,. \end{align*}\] Set \[ M:=\max \left\{\left|a_{1}\right|,\left|a_{2}\right|, \ldots,\left|a_{N-1}\right|, 1+|a|\right\} \,. \] Note that such maximum exists, being the set finite. Then \[ \left|a_{n}\right| \leq M \,, \quad \forall \, n \in \mathbb{N}\,, \] showing that \((a_n)\) is bounded.

The choice of \(M\) in the above proof says that the sequence can behave wildly for a finite number of terms. After that, it will stay close to the value of the limit, if the latter exists.

Example 20

In Theorem 8 we have shown that \[ \lim_{n \to \infty} \, \frac{1}{n} = 0 \] Hence, it follows from Theorem 136 that the sequence \((1/n)\) is bounded.

Indeed, we have that \[ \left|\frac{1}{n}\right|=\frac{1}{n} \leq 1 \,, \quad \forall \, n \in \mathbb{N}\,, \] since \(n \geq 1\) for all \(n \in \mathbb{N}\).

Warning

The converse of Theorem 136 does not hold: There exist sequences \((a_n)\) which are bounded, but not convergent.

Example 21

Define the sequence \[ a_n = (-1)^n \,. \] We have proven in Theorem 133 that \((a_n)\) is not convergent. However \((a_n)\) is bounded, with \(M=1\), since \[ |a_n| = |(-1)^n| = 1 = M \,, \quad \forall \, n \in \mathbb{N}\,. \]

Taking the contrapositive of the statement in Theorem 136 we get the following corollary:

Corollary 22

If a sequence \(\left(a_{n}\right)_{n \in \mathbb{N}}\) is not bounded, then the sequence does not converge.

Remark 23

For a sequence \((a_n)\) to be unbounded, it means that \[ \forall \, M > 0 \,, \,\, \exists \, n \in \mathbb{N}\, \text{ s.t. } \, \left|a_{n}\right| > M \,. \] The above is saying that no real number \(M>0\) can be a bound for \(|a_n|\), since there is always an index \(n \in \mathbb{N}\) such that \[ |a_n| > M \,. \]

We can use Corollary 138 to show that certain sequences do not converge.

Theorem 24

For all \(p>0\), the sequence \(\left(n^{p}\right)_{n \in \mathbb{N}}\) does not converge.

Proof

Let \(p>0\). We prove that the sequence \(\left(n^{p}\right)_{n \in \mathbb{N}}\) is unbounded, that is, \[ \forall \, M > 0 \,, \,\, \exists \, n \in \mathbb{N}\, \text{ s.t. } \, \left|a_{n}\right| > M \,. \] To this end, let \(M > 0\). Choose \(n \in \mathbb{N}\) such that \[ n > M^{1/p} \,. \] Then \[ a_n = n^{p}>\left(M^{1/p}\right)^{p}=M \,. \] This proves that the sequence \((n^p)\) is unbounded. Hence \((n^p)\) cannot converge, by Corollary 138.

Theorem 25

The sequence \((\log n)_{n \in \mathbb{N}}\) does not converge.

Proof

Let us show that \((\log n)_{n \in \mathbb{N}}\) is unbounded, that is, \[ \forall \, M > 0 \,, \,\, \exists \, n \in \mathbb{N}\, \text{ s.t. } \, \left|a_{n}\right| > M \,. \] To this end let \(M > 0\). Choose \(n \in \mathbb{N}\) such that \[ n \geq e^{M+1} \,. \] Then \[ |a_n| = |\log n| \geq \left| \log e^{M+1} \right| = M + 1 > M \,. \] This proves that the sequence \((\log n)\) is unbounded. Hence \((\log n)\) cannot converge, by Corollary 138.

6.6 Algebra of limits

Proving convergence using Definition 129 can be a tedious task. In this section we discuss how to prove convergence, starting from known convergence results.

Theorem 26: Algebra of limits

Let \(\left(a_{n}\right)_{n \in \mathbb{N}}\) and \(\left(b_{n}\right)_{n \in \mathbb{N}}\) be sequences in \(\mathbb{R}\). Suppose that \[ \lim_{n \rightarrow \infty} a_{n}=a \,, \quad \lim_{n \rightarrow \infty} b_{n}=b \,, \] for some \(a,b \in \mathbb{R}\). Then,

1. Limit of sum is the sum of limits: \[ \lim_{n \rightarrow \infty}\left(a_{n} \pm b_{n}\right)=a \pm b \]

2. Limit of product is the product of limits: \[ \lim _{n \rightarrow \infty}\left(a_{n} b_{n}\right) = a b \]

3. If \(b_{n} \neq 0\) for all \(n \in \mathbb{N}\) and \(b \neq 0\), then \[ \lim_{n \rightarrow \infty} \left(\frac{a_{n}}{b_{n}}\right)=\frac{a}{b} \]

Proof

Let \(\left(a_{n}\right)_{n \in \mathbb{N}}\) and \(\left(b_{n}\right)_{n \in \mathbb{N}}\) be sequences in \(\mathbb{R}\) and let \(c \in \mathbb{R}\). Suppose that, for some \(a, b \in \mathbb{R}\) \[ \lim _{n \rightarrow \infty} a_{n}=a \,, \quad \lim _{n \rightarrow \infty} b_{n}=b \,. \]

Proof of Point 1.

We need to show that \[ \lim_{n \to \infty} (a_n \pm b_n) = a \pm b \,. \] We only give a proof of the formula with \(+\), since the case with \(-\) follows with a very similar proof. Hence, we need to show that \[ \forall \, \varepsilon> 0 \,, \, \exists \, N \in \mathbb{N}\, \text{ s.t. } \, \forall \, n \geq N \,, \, \left|(a_{n} + b_n) - (a+b) \right| < \varepsilon\,. \] Let \(\varepsilon>0\) and set \[ \widetilde{\varepsilon} := \frac{\varepsilon}{2} \,. \] Since \(a_{n} \rightarrow a, b_n \to b\), and \(\widetilde{\varepsilon}>0\), there exist \(N_1, N_2 \in \mathbb{N}\) such that \[\begin{align*} |a_n - a| & < \widetilde{\varepsilon} \,, \quad \forall \, n \geq N_1 \,, \\ |b_n - b| & < \widetilde{\varepsilon} \,, \quad \forall \, n \geq N_2 \,. \end{align*}\] Define \[ N : = \max \{ N_1, N_2 \} \,. \] For all \(n \geq N\) we have, by the triangle inequality, \[\begin{align*} \left|\left(a_{n}+b_{n}\right)-(a+b)\right| & = \left|\left(a_{n}-a\right) + \left(b_{n}-b\right)\right| \\ & \leq \left|a_{n}-a\right|+\left|b_{n}-b\right| \\ & < \widetilde{\varepsilon}+\widetilde{\varepsilon} \\ & =\varepsilon\,. \end{align*}\]

Proof of Point 2.

We need to show that \[ \lim_{n \to \infty} (a_n b_n) = a b \,, \] which is equivalent to \[ \forall \, \varepsilon> 0 \,, \, \exists \, N \in \mathbb{N}\, \text{ s.t. } \, \forall \, n \geq N \,, \, \left|a_{n} b_{n} - a b \right| < \varepsilon\,. \] Let \(\varepsilon>0\). The sequence \(\left(a_{n}\right)\) converges, and hence is bounded, by Theorem 136. This means there exists some \(M>0\) such that \[ \left|a_{n}\right| \leq M \,, \quad \forall \, n \in \mathbb{N}\,. \] Define \[ \widetilde{\varepsilon} = \frac{\varepsilon}{ M + |b| } \,. \] Since \(a_{n} \rightarrow a, b_n \to b\), and \(\widetilde{\varepsilon} > 0\), there exist \(N_1, N_2 \in \mathbb{N}\) such that \[\begin{align*} | a_n - a | & < \widetilde{\varepsilon} \,, \quad \forall \, n \geq N_1 \,. \\ | b_n - b | & < \widetilde{\varepsilon} \,, \quad \forall \, n \geq N_2 \,. \end{align*}\] Let \[ N:= \max\{ N_1, N_2 \} \,. \] For all \(n \geq N\) we have \[\begin{align*} \left|a_{n}b_{n}-a b\right| & = \left|a_{n} b_{n} - a_n b + a_n b - a b \right| \\ & \leq \left|a_{n} b_{n} - a_n b \right| + \left|a_n b - a b \right| \\ & =\left|a_{n}\right| \left|b_{n}-b\right|+ |b| \left|a_{n}-a\right| \\ & < M \, \widetilde{\varepsilon} +|b| \, \widetilde{\varepsilon} \\ & = (M + |b|) \, \widetilde{\varepsilon} \\ & = \varepsilon\,. \end{align*}\]

Proof of Point 3.

Suppose in addition that \(b_n \neq 0\) and \(b \neq 0\). We need to show that \[ \lim_{n \to \infty} \, \frac{a_n}{b_n} = \frac{a}{b} \,, \] which is equivalent to \[ \forall \, \varepsilon> 0 \,, \, \exists \, N \in \mathbb{N}\, \text{ s.t. } \, \forall \, n \geq N \,, \, \left|\frac{a_{n}}{ b_{n}} - \frac{a}{b} \right| < \varepsilon\,. \] We suppose in addition that \(b>0\). The proof is very similar for the case \(b <0\), and is hence omitted. Let \(\varepsilon> 0\). Set \[ \delta := \frac{b}{2} \,. \] Since \(b_n \to b\) and \(\delta>0\), there exists \(N_1 \in \mathbb{N}\) such that \[ |b_n - b| < \delta \, \quad \forall \, n \geq N_1 \,. \] In particular we have \[ b_n > b - \delta = b - \frac{b}{2} = \frac{b}{2} \, \quad \forall \, n \geq N_1 \,. \] Define \[ \widetilde{\varepsilon} := \frac{ b^2 }{ 2 (b + |a|)} \, \varepsilon\,. \] Since \(\widetilde{\varepsilon}>0\) and \(a_n \to a, b_n \to b\), there exist \(N_2, N_3 \in \mathbb{N}\) such that \[\begin{align*} |a_n - a | & < \widetilde{\varepsilon} \,, \quad \forall \, n \geq N_2 \,, \\ |b_n - b | & < \widetilde{\varepsilon} \,, \quad \forall \, n \geq N_3 \,. \end{align*}\] Define \[ N:= \max\{ N_1, N_2, N_3 \} \,. \] For all \(n \geq N\) we have \[\begin{align*} \left|\frac{a_{n}}{ b_{n}} - \frac{a}{b} \right| & = \left|\frac{a_{n} b - a b_n}{ b_{n} b } \right| \\ & = \frac{1}{ \left|b_{n} b \right|} \, \left| a_{n}b - a b + a b - a b_n \right| \\ & = \frac{1}{ \left|b_{n} b \right|} \, \left| (a_{n} - a) b + a(b-b_n) \right| \\ & \leq \frac{1}{ \left|b_{n} b \right|} \, \left( |a_{n} - a| |b| + |a| |b-b_n| \right) \\ & < \frac{1}{ \dfrac{b}{2} \, b } \, \left( \widetilde{\varepsilon} \, b + \widetilde{\varepsilon} |a| \right) \\ & = \frac{2 (b + |a|) }{b^2} \, \widetilde{\varepsilon} \\ & = \varepsilon\,. \end{align*}\]

In the future we will refer to Theorem 143 as the Algebra of Limits. We now show how to use Theorem 143 for computing certain limits.

Example 27

Question. Prove that \[ \lim_{n \to \infty} \, \frac{3 n}{7 n+4} = \frac{3}{7} \,. \]

Solution. We can rewrite \[ \frac{3 n}{7 n+4}=\frac{3}{7+\dfrac{4}{n}} \] By Theorem 132 we know that \[ 3 \rightarrow 3\,, \quad 4 \to 4 \,, \quad 7 \to 7 \,. \] From Theorem 8 we know that \[ \frac{1}{n} \rightarrow 0 \,. \] Hence, it follows from Theorem 143 Point 2 that \[ \frac{4}{n} = 4 \cdot \frac1n \rightarrow 4 \cdot 0 = 0 \,. \] By Theorem 143 Point 1 we have \[ 7 + \frac{4}{n} \rightarrow 7 + 0 = 7 \,. \] Finally we can use Theorem 143 Point 3 to infer \[ \frac{3 n}{7 n+4}=\frac{3}{7+\dfrac{4}{n}} \rightarrow \frac{3}{7} \,. \]

Important

The technique shown in Example 143 is useful to compute limits of fractions of polynomials. To identify the possible limit, if it exists, it is often best to divide by the largest power of \(n\) in the denominator.

Example 28

Question. Prove that \[ \lim_{n \rightarrow \infty} \frac{n^{2}-1}{2 n^{2}-3} = \frac{1}{2} \,. \]

Solution. Factor \(n^2\) to obtain \[ \frac{n^{2}-1}{2 n^{2}-3} = \frac{1-\dfrac{1}{n^{2}}}{2-\dfrac{3}{n^{2}}} \,. \] By Theorem 130 we have \[ \frac{1}{n^2} \to 0 \,. \] We can then use the Algebra of Limits Theorem 143 Point 2 to infer \[ \frac{3}{n^2} \to 3 \cdot 0 = 0 \] and Theorem 143 Point 1 to get \[ 1 - \frac{1}{n^2} \to 1 - 0 = 1 \,, \quad 2 - \frac{3}{n^2} \to 2 - 0 = 2 \,. \] Finally we use Theorem 143 Point 3 and conclude \[ \frac{1-\dfrac{1}{n^{2}}}{2-\dfrac{3}{n^{2}}} \to \frac{1}{2} \,. \] Therefore \[ \lim_{n \to \infty } \, \frac{n^{2}-1}{2 n^{2}-3} = \lim_{n \to \infty} \, \frac{1-\dfrac{1}{n^{2}}}{2-\dfrac{3}{n^{2}}} = \frac{1}{2} \,. \]

We can also use the Algebra of Limits to prove that certain limits do not exist.

Example 29

Question. Prove that the sequence \[ a_n = \frac{4 n^{3}+8 n+1}{7 n^{2}+2 n+1} \] does not converge.

Solution. To show that the sequence \(\left(a_n\right)\) does not converge, we divide by the largest power in the denominator, which in this case is \(n^2\) \[\begin{align*} a_n & = \frac{4 n^{3}+8 n+1}{7 n^{2}+2 n+1} \\ & =\frac{4 n+ \dfrac{8}{n} + \dfrac{1}{n^{2}} }{7+ \dfrac{2}{n} + \dfrac{1}{n^{2}} } \\ & = \frac{b_n}{c_n} \end{align*}\] where we set \[ b_n := 4 n+ \dfrac{8}{n} + \dfrac{1}{n^{2}}\,, \quad c_n := 7 + \dfrac{2}{n} + \dfrac{1}{n^{2}} \,. \] Using the Algebra of Limits Theorem 143 we see that \[ c_n = 7+ \dfrac{2}{n} + \dfrac{1}{n^{2}} \to 7 \,. \] Suppose by contradiction that \[ a_n \to a \] for some \(a \in \mathbb{R}\). Then, by the Algebra of Limits Theorem 143 we would infer \[ b_n = c_n \cdot a_n \to 7 a \,, \] concluding that \(b_n\) is convergent to \(7a\). We have that \[ b_n = 4n + d_n \,, \quad d_n := \dfrac{8}{n} + \dfrac{1}{n^{2}} \,. \] Again by the Algebra of Limits Theorem 143 we get that \[ d_n = \dfrac{8}{n} + \dfrac{1}{n^{2}} \to 0 \,, \] and hence \[ 4n = b_n - d_n \to 7a - 0 = 7a \,. \] This is a contradiction, since the sequence \((4n)\) is unbounded, and hence cannot be convergent. Hence \((a_n)\) is not convergent.

Warning

Consider the sequence \[ a_n = \frac{4 n^{3}+8 n+1}{7 n^{2}+2 n+1} \] from the previous example. We have proven that \((a_n)\) is not convergent, by making use of the Algebra of Limits.

Let us review a faulty argument to conclude that \((a_n)\) is not convergent. Write \[ a_n = \frac{b_n}{c_n} \,, \quad b_n:= 4 n^{3}+8 n+1\,, \quad c_n : =7 n^{2}+2 n+1 \,. \] The numerator \[ b_n = 4 n^{3}+8 n+1 \] and denominator \[ c_n =7 n^{2}+2 n+1 \] are both unbounded, and hence \((b_n)\) and \((c_n)\) do not converge. One might be tempted to conclude that \((a_n)\) does not converge. However this is false in general: as seen in Example 144, we have \[ \lim_{n \rightarrow \infty} \frac{n^{2}-1}{2 n^{2}-3} = \frac{1}{2} \,, \] while numerator and denominator are unbounded.

Sometimes it is useful to rearrange the terms of a sequence, before applying the Algebra of Limits.

Example 30

Question. Define the sequence \[ a_n := \frac{2 n^{3}+7 n+1}{5 n+9} \cdot \frac{8 n+9}{6 n^{3}+8 n^{2}+3} \,. \] Prove that \[ \lim_{n \to \infty} a_n = \frac{8}{15} \,. \]

Solution. The first fraction in \((a_n)\) does not converge, as it is unbounded. Therefore we cannot use Point 2 in Theorem 143 directly. However, we note that \[\begin{align*} a_{n} & =\frac{2 n^{3}+7 n+1}{5 n+9} \cdot \frac{8 n+9}{6 n^{3}+8 n^{2}+3} \\ & = \frac{8 n+9}{5 n+9} \cdot \frac{2 n^{3}+7 n+1}{6 n^{3}+8 n^{2}+3} \,. \end{align*}\] Factoring out \(n\) and \(n^3\), respectively, and using the Algebra of Limits, we see that \[ \frac{8 n+9}{5 n+9}=\frac{8+9 / n}{5+9 / n} \to \frac{8+0}{5+0}=\frac{8}{5} \] and \[ \frac{2+7 / n^{2}+1 / n^{3}}{6+8 / n+3 / n^{3}} \to \frac{2+0+0}{6+0+0}=\frac{1}{3} \] Therefore Theorem 143 Point 2 ensures that \[ a_{n} \to \frac{8}{5} \cdot \frac{1}{3}=\frac{8}{15} \,. \]

6.7 Fractional powers

The Algebra of Limits Theorem 143 can also be used when fractional powers of \(n\) are involved.

Example 31

Question. Prove that \[ a_n = \frac{n^{7 / 3}+2 \sqrt{n}+7}{4 n^{3 / 2}+5 n} \] does not converge.

Solution. The largest power of \(n\) in the denominator is \(n^{3/2}\). Hence we factor out \(n^{3/2}\) \[\begin{align*} a_n & = \frac{n^{7 / 3}+2 \sqrt{n}+7}{4 n^{3 / 2}+5 n} \\ & = \frac{n^{7 / 3-3 / 2}+2 n^{1/2 - 3/2} + 7 n^{-3 / 2}}{4 + 5 n^{-3/2} } \\ & = \frac{n^{5/6}+2 n^{- 1} + 7 n^{-3 / 2}}{4 + 5 n^{-3/2} } \\ & = \frac{b_n}{c_n} \end{align*}\] where we set \[ b_n := n^{5/6}+2 n^{- 1} + 7 n^{-3 / 2} \,, \quad c_n := 4 + 5 n^{-3/2} \,. \] We see that \(b_n\) is unbounded while \(c_n \to 4\). By the Algebra of Limits (and usual contradiction argument) we conclude that \((a_n)\) is divergent.

We now present a general result about the square root of a sequence.

Theorem 32

Let \(\left(a_{n}\right)_{n \in \mathbb{N}}\) be a sequence in \(\mathbb{R}\) such that \[ \lim_{n \to \infty} \, a_n = a \,, \] for some \(a \in \mathbb{R}\). If \(a_{n} \geq 0\) for all \(n \in \mathbb{N}\) and \(a \geq 0\), then \[ \lim _{n \rightarrow \infty} \sqrt{a_{n}}=\sqrt{a} \,. \]

Proof

Let \(\varepsilon>0\). We the two cases \(a>0\) and \(a=0\):

• \(a>0\): Define \[ \delta := \frac{a}{2} \,. \] Since \(\delta > 0\) and \(a_n \to a\), there exists \(N_1 \in \mathbb{N}\) such that \[ \left|a_{n}-a\right| < \delta \,, \quad \forall \, n \geq N_1 \,. \] In particular \[ a_n > a - \delta = a - \frac{a}{2} = \frac{a}{2} \,, \quad \forall \, n \geq N_1 \,, \] from which we infer \[ \sqrt{a_n} > \sqrt{a/2} \,, \quad \forall \, n \geq N_1 \,, \] Now set \[ \widetilde{\varepsilon} := \left(\sqrt{a/2} + \sqrt{a} \right) \, \varepsilon\,. \] Since \(\widetilde{\varepsilon}>0\) and \(a_n \to a\), there exists \(N_2 \in \mathbb{N}\) such that \[ \left|a_{n}-a\right| < \widetilde{\varepsilon} \,, \quad \forall \, n \geq N_2 \,. \] Let \[ N:= \max\{ N_1, N_2 \} \,. \] For \(n \geq N\) we have \[\begin{align*} \left|\sqrt{a_{n}}-\sqrt{a}\right| & = \left|\frac{\left(\sqrt{a_{n}}-\sqrt{a}\right)\left(\sqrt{a_{n}}+\sqrt{a}\right)}{\sqrt{a_{n}}+\sqrt{a}}\right| \\ & = \frac{\left|a_{n}-a\right|}{\sqrt{a_{n}}+\sqrt{a}} \\ & < \frac{ \widetilde{\varepsilon} }{\sqrt{a/2} + \sqrt{a}} \\ & = \varepsilon\,. \end{align*}\]

• \(a=0\): In this case \[ a_n \to a = 0 \,. \] Since \(\varepsilon^2>0\), there exists \(N \in \mathbb{N}\) such that \[ |a_n - 0 | = |a_n| < \varepsilon^2 \,, \quad \forall \, n \geq N \,. \] Therefore \[ |\sqrt{a_n} - \sqrt{0} | = | \sqrt{a_n}| < \sqrt{\varepsilon^2} = \varepsilon\,, \quad \forall \, n \geq N \,. \]

Let us show an application of Theorem 148.

Example 33

Question. Define the sequence \[ a_n = \sqrt{9 n^{2}+3 n+1}-3 n \,. \] Prove that \[ \lim_{n \to \infty} \, a_n = \frac12 \,. \]

Solution. We first rewrite \[\begin{align*} a_n & = \sqrt{9 n^{2}+3 n+1}-3 n \\ & = \frac{\left(\sqrt{9 n^{2}+3 n+1}-3 n\right)\left(\sqrt{9 n^{2}+3 n+1}+3 n\right)}{\sqrt{9 n^{2}+3 n+1}+3 n} \\ & =\frac{9 n^{2}+3 n+1-(3 n)^{2}}{\sqrt{9 n^{2}+3 n+1}+3 n} \\ & =\frac{3 n+1}{\sqrt{9 n^{2}+3 n+1}+3 n} \, . \end{align*}\] The biggest power of \(n\) in the denominator is \(n\). Therefore we factor out \(n\): \[\begin{align*} a_n & = \sqrt{9 n^{2}+3 n+1}-3 n \\ & =\frac{3 n+1}{\sqrt{9 n^{2}+3 n+1}+3 n} \\ & =\frac{3+ \dfrac{1}{n} }{\sqrt{9+ \dfrac{3}{n} + \dfrac{1}{n^{2}}} + 3 } \,. \end{align*}\] By the Algebra of Limits we have \[ 9+ \frac{3}{n} + \frac{1}{n^{2}} \to 9 + 0 + 0 = 9 \,. \] Therefore we can use Theorem 148 to infer \[ \sqrt{ 9 + \frac{3}{n} + \frac{1}{n^{2}} } \to \sqrt{9} \,. \] By the Algebra of Limits we conclude: \[ a_n = \frac{3+ \dfrac{1}{n} }{\sqrt{9+ \dfrac{3}{n} + \dfrac{1}{n^{2}} }+ 3 } \to \frac{ 3 + 0 }{ \sqrt{9} + 3 } = \frac12 \,. \]

Example 34

Question. Prove that the sequence \[ a_n = \sqrt{9 n^{2}+3 n+1}-2 n \] does not converge.

Solution. We rewrite \(a_n\) as \[\begin{align*} a_n & = \sqrt{9 n^{2}+3 n+1}-2 n \\ & =\frac{ (\sqrt{9 n^{2}+3 n+1} - 2 n) (\sqrt{9 n^{2}+3 n+1}+2 n) }{\sqrt{9 n^{2}+3 n+1}+2 n} \\ & =\frac{9 n^{2}+3 n+1-(2 n)^{2}}{\sqrt{9 n^{2}+3 n+1}+2 n} \\ & =\frac{5 n^{2}+3 n+1}{\sqrt{9 n^{2}+3 n+1}+2 n} \\ & =\frac{5 n + 3 + \dfrac{1}{n} }{\sqrt{9+ \dfrac{3}{n} + \dfrac{1}{ n^2 } } + 2 } \\ & = \frac{b_n}{c_n} \,, \end{align*}\] where we factored \(n\), being it the largest power of \(n\) in the denominator, and defined \[ b_n : = 5 n + 3 + \dfrac{1}{n}\,, \quad c_n := \sqrt{9+ \dfrac{3}{n} + \dfrac{1}{ n^2 } } + 2 \,. \] Note that \[ 9+ \dfrac{3}{n} + \dfrac{1}{ n^2 } \to 9 \] by the Algebra of Limits. Therefore \[ \sqrt{9+ \dfrac{3}{n} + \dfrac{1}{ n^2 }} \to \sqrt{9} = 3 \] by Theorem 148. Hence \[ c_n = \sqrt{9+ \dfrac{3}{n} + \dfrac{1}{ n^2 }} + 2 \to 3 + 2 = 5 \,. \] The numerator \[ b_n = 5 n + 3 + \dfrac{1}{n} \] is instead unbounded. Therefore \((a_n)\) is not convergent, by the Algebra of Limits and the usual contradiction argument.

6.8 Limit Tests

In this section we discuss a number of Tests to determine whether a sequence converges or not. These are known as Limit Tests.

6.8.1 Squeeze Theorem

When a sequence \((a_n)\) oscillates, it is difficult to compute the limit. Examples of terms which produce oscillations are \[ (-1)^n \,, \quad \sin(n) \,, \quad \cos(n) \,. \] In such instance it might be useful to compare \((a_n)\) with other sequences whose limit is known. If we can prove that \((a_n)\) is squeezed between two other sequences with the same limiting value, then we can show that also \((a_n)\) converges to this value.

Theorem 35: Squeeze theorem

Let \(\left(a_{n}\right), \left(b_{n}\right)\) and \(\left(c_{n}\right)\) be sequences in \(\mathbb{R}\). Suppose that \[ b_{n} \leq a_{n} \leq c_{n} \,, \quad \forall \, n \in \mathbb{N}\,, \] and that \[ \lim_{n \rightarrow \infty} b_{n} = \lim_{n \rightarrow \infty} c_{n} = L \, . \] Then \[ \lim_{n \rightarrow \infty} a_{n}= L \, . \]

Proof

Let \(\varepsilon>0\). Since \(b_{n} \to L\) and \(c_n \to L\) , there exist \(N_1, N_2 \in \mathbb{N}\) such that \[\begin{align*} -\varepsilon< b_n - L < \varepsilon\,, \,\, \forall \, n \geq N_1 \,, \\ - \varepsilon< c_n - L < \varepsilon\,, \,\, \forall \, n \geq N_2 \,. \end{align*}\] Set \[ N := \max\{N_1,N_2\} \,. \] Let \(n \geq N\). Using the assumption that \(b_n \leq a_{n} \leq c_{n}\), we get \[ b_n - L \leq a_{n} - L \leq c_{n} - L \,. \] In particular \[ - \varepsilon< b_n - L \leq a_n - L \leq b_n - L < \varepsilon\,. \] The above implies \[ - \varepsilon< a_n - L < \varepsilon\quad \implies \quad \left|a_{n}-L\right| < \varepsilon\,. \]

Example 36

Question. Prove that \[ \lim_{n \to \infty} \, \frac{(-1)^{n}}{n} = 0 \,. \]

Solution. For all \(n \in \mathbb{N}\) we can estimate \[ -1 \leq(-1)^{n} \leq 1 \,. \] Therefore \[ \frac{-1}{n} \leq \frac{(-1)^{n}}{n} \leq \frac{1}{n} \,, \quad \forall \, n \in \mathbb{N}\,. \] Moreover \[ \lim_{n \to \infty} \frac{-1}{n}= -1 \cdot 0=0 \,, \quad \lim_{n \to \infty} \frac{1}{n}=0 \,. \] By the Squeeze Theorem 151 we conclude \[ \lim_{n \to \infty} \frac{(-1)^{n}}{n}=0 \,. \]

Example 37

Question. Prove that \[ \lim_{n \to \infty} \frac{\cos (3 n) + 9 n^{2}}{11 n^{2}+15 \sin (17 n)} = \frac{9}{11} \,. \]

Solution. We know that \[ -1 \leq \cos(x) \leq 1 \,, \quad - 1 \leq \sin(x) \leq 1 \,, \quad \forall \, x \in \mathbb{R}\,. \] Therefore, for all \(n \in \mathbb{N}\) \[ - 1 \leq \cos(3n) \leq 1 \,, \quad -1 \leq \sin(17n) \leq 1 \,. \] We can use the above to estimate the numerator in the given sequence: \[ -1 + 9 n^{2} \leq \cos (3 n)+9 n^{2} \leq 1+ 9 n^{2} \,. \tag{6.7}\] Concerning the denominator, we have \[ 11 n^{2}-15 \leq 11 n^{2}+15 \sin (17 n) \leq 11 n^{2} + 15 \] and therefore \[ \frac{1}{11 n^{2} + 15} \leq \frac{1}{11 n^{2}+15 \sin (17 n)} \leq \frac{1}{11 n^{2}-15} \,. \tag{6.8}\] Putting together (A.31)-(A.32) we obtain \[ \frac{-1 + 9 n^{2}}{11 n^{2} + 15} \leq \frac{\cos (3 n)+9 n^{2}}{11 n^{2}+15 \sin (17 n)} \leq \frac{1+ 9 n^{2}}{11 n^{2}-15} \,. \] By the Algebra of Limits we infer \[ \frac{-1+9 n^{2}}{11 n^{2}+15}=\frac{-\dfrac{1}{n^{2}} + 9}{11 + \dfrac{15}{n^{2}}} \to \frac{0+9}{11+0}=\frac{9}{11} \] and \[ \frac{1+9 n^{2}}{11 n^{2} - 15}=\frac{ \dfrac{1}{n^{2}} + 9}{ 11 - \dfrac{15}{n^{2}}} \to \frac{0+9}{11+0}=\frac{9}{11} \,. \] Applying the Squeeze Theorem 151 we conclude \[ \lim_{n \to \infty} \frac{\cos (3 n)+ 9 n^{2}}{11 n^{2}+15 \sin (17 n)}=\frac{9}{11} \,. \]

Warning

Suppose that the sequences \((a_n), (b_n), (c_n)\) satisfy \[ b_n \leq a_n \leq c_n \,, \quad \forall n \in \mathbb{N}\,, \] and \[ b_n \to L_1 \,, \quad c_n \to L_2 \,, \quad L_1 \neq L_2 \,. \] In general, we cannot conclude that \(a_n\) converges.

Example 38

Consider the sequence \[ a_n = \left( 1 + \frac{1}{n} \right) (-1)^n \,. \] For all \(n \in \mathbb{N}\) we can bound \[ - 1 - \frac{1}{n} \leq \left( 1 + \frac{1}{n} \right) (-1)^n \leq 1 + \frac{1}{n} \,. \] However \[ - 1 - \frac{1}{n} \longrightarrow - 1 - 0 = -1 \] and \[ 1 + \frac{1}{n} \longrightarrow 1 + 0 = 1 \,. \] Since \[ - 1 \neq 1 \,, \] we cannot apply the Squeeze Theorem 151 to conclude convergence of \((a_n)\). Indeed, \((a_n)\) is a divergent sequence.

Proof. Suppose by contradiction that \(a_n \to a\). We have \[ a_n = (-1)^n + \frac{(-1)^n}{n} = b_n + c_n \] where \[ b_n := (-1)^n \,, \quad c_n := \frac{(-1)^n}{n} \,. \] We have seen in Example 153 that \(c_n \to 0\). Therefore, by the Algebra of Limits, we have \[ b_n = a_n - c_n \longrightarrow a - 0 = a \,. \] However, Theorem 133 says that the sequence \(b_n = (-1)^n\) diverges. Contradiction. Hence \((a_n)\) diverges.

6.8.2 Geometric sequences

Definition 39

A sequence \(\left(a_{n}\right)\) is called a geometric sequence if \[ a_{n}=x^{n} \,, \] for some \(x \in \mathbb{R}\).

The value of \(|x|\) determines whether or not a geometric sequence converges, as shown in the following theorem.

Theorem 40: Geometric Sequence Test

Let \(x \in \mathbb{R}\) and let \(\left(a_{n}\right)\) be the sequence defined by \[ a_{n}:=x^{n} \,. \] We have:

1. If \(|x|<1\), then \[ \lim_{n \to \infty} a_{n} = 0 \,. \]

2. If \(|x|>1\), then sequence \(\left(a_{n}\right)\) is unbounded, and hence divergent.

Warning

The Geometric Sequence Test in Theorem 156 does not address the case \[ |x|=1 \,. \] This is because, in this case, the sequence \[ a_n = x^n \] might converge or diverge, depending on the value of \(x\). Indeed, \[ |x| = 1 \quad \implies \quad x = \pm 1 \,. \] We therefore have two cases:

• \(x = 1\): Then \[ a_n = 1^n = 1 \] so that \(a_n \to 1\) and \((a_n)\) is convergent.

• \(x=-1\): Then \[ a_n = x^n = (-1)^n \] which is divergent by Theorem 133.

To prove Theorem 156 we need the following inequality, known as Bernoulli’s inequality.

Lemma 41: Bernoulli’s inequality

Let \(x \in \mathbb{R}\) with \(x>-1\). Then \[ (1+x)^{n} \geq 1+n x \,, \quad \forall \, n \in \mathbb{N}\,. \tag{6.9}\]

Proof

Let \(x \in \mathbb{R}, x>-1\). We prove the statement by induction:

• Base case: (A.33) holds with equality when \(n=1\).

• Induction hypothesis: Let \(k \in \mathbb{N}\) and suppose that (A.33) holds for \(n=k\), i.e., \[ (1+x)^{k} \geq 1+k x \,. \] Then \[\begin{align*} (1+x)^{k+1} & = (1+x)^{k}(1+x) \\ & \geq(1+k x)(1+x) \\ & =1+k x+x+k x^{2} \\ & \geq 1+(k+1) x \,, \end{align*}\] where we used that \(kx^2 \geq 0\). Then (A.33) holds for \(n=k+1\).

By induction we conclude (A.33).

We are ready to prove Theorem 156.

Proof: Proof of Theorem 156

Part 1. The case \(|x|<1\).

If \(x=0\), then \[ a_n = x^n = 0 \] so that \(a_n \to 0\). Hence assume \(x \neq 0\). We need to prove that \[ \forall \, \varepsilon> 0 \,, \, \exists \, N \in \mathbb{N}\, \text{ s.t. } \, \forall \, n \geq N \,, \,\, |x^n - 0| < \varepsilon\,. \] Let \(\varepsilon>0\). We have \[ |x| < 1 \quad \implies \quad \frac{1}{|x|} > 1 \,. \] Therefore \[ |x|= \frac{1}{1+u} \,, \quad u:=\frac{1}{|x|} - 1 > 0 \,. \] Let \(N \in \mathbb{N}\) be such that \[ N > \frac{1}{\varepsilon u} \,, \] so that \[ \frac{1}{N u} < \varepsilon\,. \] Let \(n \geq N\). Then \[\begin{align*} \left|x^{n}-0\right| & =|x|^{n} \\ & = \left(\frac{1}{1+u}\right)^{n} \\ & =\frac{1}{(1+u)^{n}} \\ & \leq \frac{1}{1+n u} \\ & \leq \frac{1}{n u} \\ & \leq \frac{1}{N u} \\ & < \varepsilon\,, \end{align*}\] where we used Bernoulli’s inequality (A.33) in the first inequality.

Part 2. The case \(|x|>1\).

To prove that \((a_n)\) does not converge, we prove that it is unbounded. This means showing that \[ \forall \, M > 0 \,, \, \exists n \in \mathbb{N}\, \text{ s.t. } \, \left| a_{n}\right| >M \,. \] Let \(M > 0\). We have two cases:

• \(0 < M \leq 1\): Choose \(n=1\). Then \[ \left|a_{1}\right|=|x|>1 \geq M \,. \]

• \(M>1\): Choose \(n \in \mathbb{N}\) such that \[ n> \frac{\log M}{\log |x|} \,. \] Note that \(\log |x|>0\) since \(|x|>1\). Therefore \[\begin{align*} n>\frac{\log M}{\log |x|} & \iff n \log |x|>\log M \\ & \iff \log |x|^n>\log M \\ & \iff |x|^{n}>M \,. \end{align*}\] Then \[ \left|a_{n}\right|=\left|x^{n}\right|=|x|^{n} > M \,. \]

Hence \((a_n)\) is unbounded. By Corollary 138 we conclude that \((a_n)\) is divergent.

Example 42

We can apply Theorem 156 to prove convergence or divergence for the following sequences.

1. We have \[ \left(\frac{1}{2}\right)^{n} \longrightarrow 0 \] since \[ \left|\frac{1}{2}\right|=\frac{1}{2}<1 \,. \]

2. We have \[ \left(\frac{-1}{2}\right)^{n} \longrightarrow 0 \] since \[ \left|\frac{-1}{2}\right|=\frac{1}{2}<1 \,. \]

3. The sequence \[ a_n = \left(\frac{-3}{2}\right)^{n} \] does not converge, since \[ \left|\frac{-3}{2}\right|=\frac{3}{2}>1 \,. \]

4. As \(n \rightarrow \infty\), \[ \frac{3^{n}}{(-5)^{n}}=\left(-\frac{3}{5}\right)^{n} \longrightarrow 0 \] since \[ \left|-\frac{3}{5}\right|=\frac{3}{5}<1 \,. \]

5. The sequence \[ a_{n}=\frac{(-7)^{n}}{2^{2 n}} \] does not converge, since \[ \frac{(-7)^{n}}{2^{2 n}}=\frac{(-7)^{n}}{\left(2^{2}\right)^{n}}=\left(-\frac{7}{4}\right)^{n} \] and \[ \left|-\frac{7}{4}\right|=\frac{7}{4}>1 \,. \]

6.8.3 Ratio Test

Theorem 43: Ratio Test

Let \(\left(a_{n}\right)\) be a sequence in \(\mathbb{R}\) such that \[ a_{n} \neq 0 \,, \quad \forall \, n \in \mathbb{N}\,. \]

1. Suppose that the following limit exists: \[ L:=\lim_{n \to \infty}\left|\frac{a_{n+1}}{a_{n}}\right| \,. \] Then,

   • If \(L<1\) we have \[ \lim_{n \to\infty} a_{n}=0 \,. \]

   • If \(L>1\), the sequence \(\left(a_{n}\right)\) is unbounded, and hence does not converge.

2. Suppose that there exists \(N \in \mathbb{N}\) and \(L>1\) such that \[ \left|\frac{a_{n+1}}{a_{n}}\right| \geq L \,, \quad \forall \, n \geq N \,. \] Then the sequence \(\left(a_{n}\right)\) is unbounded, and hence does not converge.

Proof

Define the sequence \(b_{n}=\left|a_{n}\right|\). Then,

\[ \left|\frac{a_{n+1}}{a_{n}}\right|=\frac{\left|a_{n+1}\right|}{\left|a_{n}\right|}=\frac{b_{n+1}}{b_{n}} \]

Part 1. Suppose that there exists the limit \[ L:=\lim_{n \to \infty}\left|\frac{a_{n+1}}{a_{n}}\right| \,. \] Therefore \[ \lim_{n \to \infty} \frac{b_{n+1}}{b_{n}} = L \,. \tag{6.10}\]

• \(L<1\): Choose \(r>0\) such that \[ L<r<1 \,. \] Set \[ \varepsilon:= r-L \] By the convergence at (A.34) there exists \(N \in \mathbb{N}\) such that \[ \left|\frac{b_{n+1}}{b_{n}}-L\right| < \varepsilon= r - L \,, \quad \forall \, n \geq N \,. \] In particular \[ \frac{b_{n+1}}{b_{n}}-L < r-L \,, \quad \forall \, n \geq N \,, \] which implies \[ b_{n+1} < r \,{b_{n}} \,, \quad \forall \, n \geq N \,. \tag{6.11}\] Let \(n \geq N\), we can use (A.35) recursively and obtain \[ 0 \leq b_{n} < \, r b_{n-1} < \ldots < r^{n-N} \, b_{N} = r^{n} \, \frac{b_{N}}{r^{N}} \,. \] In particular, we have proven that \[ 0 \leq b_{n} < r^{n} \, \frac{b_{N}}{r^{N}} \,, \quad \forall \, n \in \mathbb{N}\,. \tag{6.12}\] Since \(|r|<1\), by the Geometric Sequence Test Theorem 156 we infer \[ r^{n} \rightarrow 0 \,. \] The Algebra of Limits the yields \[ r^{n} \, \frac{b_{N}}{r^{N}} \rightarrow 0 \,. \] By the Squeeze Theorem 151 applied to (A.36), it follows that \[ b_{n} = |a_n| \to 0 \,. \] Since \[ -\left|a_{n}\right| \leq a_{n} \leq\left|a_{n}\right| \,, \] and \[ -|a_n| \to 0 \,, \quad |a_n| \to 0 \,, \] we can again apply the Squeeze Theorem 151 to infer \[ a_{n} \rightarrow 0 \,. \]

• \(L>1\): Choose \(r>0\) such that \[ 1<r<L \,. \] Define \[ \varepsilon:= L - r > 0 \,. \] By the convergence (A.34), there exists \(N \in \mathbb{N}\) such that \[ \left|\frac{b_{n+1}}{b_{n}}-L\right| < \varepsilon= L - r \,, \quad \forall \, n \geq N \,. \] In particular, \[ -(L-r) < \frac{b_{n+1}}{b_{n}}-L \,, \quad \forall \, n \geq N \,, \] which implies \[ b_{n+1} > r \, b_{n} \,, \quad \forall \, n \geq N \,. \tag{6.13}\] Let \(n \geq N\). Applying (A.37) recursively we get \[ b_{n} > r^{n-N} \, b_{N} = r^{n} \, \frac{b_{N}}{r^{N}} \,, \quad \forall \, n \geq N \,. \tag{6.14}\] Since \(|r|>1\), by the Geometric Sequence Test we have that the sequence \[ (r^n) \] is unbounded. Therefore also the right hand side of (A.38) is unbounded, proving that \((b_n)\) is unbounded. Since \[ b_n = |a_n|\,, \] we conclude that \((a_n)\) is unbounded. By Corollary 138 we conclude that \((a_n)\) does not converge.

Part 2. Suppose that there exists \(N \in \mathbb{N}\) and \(L>1\) such that \[ \left|\frac{a_{n+1}}{a_{n}}\right| \geq L \,, \quad \forall \, n \geq N \,. \] Since \(b_n = |a_n|\), we infer \[ b_{n+1} \geq L \, b_n \,, \quad \forall \, n \geq N \,. \] Arguing as above, we obtain \[ b_n \geq L^n \, \frac{b_N}{L^N} \,, \quad \forall \, n \geq N \,. \] Since \(L>1\), we have that the sequence \[ L^n \, \frac{b_N}{L^N} \] is unbounded, by the Geometric Sequence Test. Hence also \((b_n)\) is unbounded, from which we conclude that \((a_n)\) is unbounded. By Corollary 138 we conclude that \((a_n)\) does not converge.

Let us apply the Ratio Test to some concrete examples.

Example 44

Question. Let \[ a_{n}=\frac{3^{n}}{n !} \,, \] where we recall that \(n!\) (pronounced \(n\) factorial) is defined by \[ n! := n \cdot (n-1) \cdot (n-2) \cdot \ldots \cdot 3 \cdot 2 \cdot 1 \,. \] Prove that \[ \lim_{n \to \infty} a_n = 0 \,. \]

Solution. We have \[\begin{align*} \left|\frac{a_{n+1}}{a_{n}}\right| & = \dfrac{\left( \dfrac{3^{n+1}}{(n+1) !} \right) }{ \left( \dfrac{3^{n}}{n !} \right) } \\ & = \frac{3^{n+1}}{3^{n}} \, \frac{n !}{(n+1) !} \\ & = \frac{3 \cdot 3^n}{3^n} \, \frac{n!}{(n+1) n!} \\ & =\frac{3}{n+1} \longrightarrow L = 0 \,. \end{align*}\] Hence, \(L=0<1\) so \(a_{n} \to 0\) by the Ratio Test in Theorem 159.

Example 45

Question. Consider the sequence \[ a_{n}=\frac{n ! \cdot 3^{n}}{\sqrt{(2 n) !}} \,. \] Prove that \((a_n)\) is divergent.

Solution. We have \[\begin{align*} \left|\frac{a_{n+1}}{a_{n}}\right| & =\frac{(n+1) ! \cdot 3^{n+1}}{\sqrt{(2(n+1)) !}} \frac{\sqrt{(2 n) !}}{n ! \cdot 3^{n}} \\ & =\frac{(n+1) !}{n !} \cdot \frac{3^{n+1}}{3^{n}} \cdot \frac{\sqrt{(2 n) !}}{\sqrt{(2(n+1)) !}} \end{align*}\] For the first two fractions we have \[ \frac{(n+1) !}{n !} \cdot \frac{3^{n+1}}{3^{n}} = 3(n+1) \,, \] while for the third fraction \[\begin{align*} \frac{\sqrt{(2 n) !}}{\sqrt{(2(n+1)) !}} & =\sqrt{\frac{(2 n) !}{(2 n+2) !}} \\ & = \sqrt{\frac{ (2n)! }{ (2n+2) \cdot (2n+1) \cdot (2n)! }} \\ & = \frac{1}{\sqrt{(2 n+1)(2 n+2)}} \,. \end{align*}\] Therefore, using the Algebra of Limits, \[\begin{align*} \left|\frac{a_{n+1}}{a_{n}}\right| & = \frac{3(n+1)}{\sqrt{(2 n+1)(2 n+2)}}\\ & = \frac{3n \left(1+ \dfrac{1}{n} \right)}{\sqrt{n^2 \left( 2 + \dfrac{1}{n}\right) \left(2 + \dfrac{2}{n} \right)}} \\ & = \frac{3 \left(1+ \dfrac{1}{n} \right)}{\sqrt{ \left( 2 + \dfrac{1}{n}\right) \left(2 + \dfrac{2}{n} \right)}} \longrightarrow \frac{3}{\sqrt{4}} = \frac{3}{2} > 1 \,. \end{align*}\] By the Ratio Test we conclude that \((a_n)\) is divergent.

Example 46

Question. Prove that the following sequence is divergent \[ a_{n}=\frac{n !}{100^{n}} \,. \]

Solution. We have \[ \left|\frac{a_{n+1}}{a_{n}}\right| = \frac{100^{n}}{100^{n+1}} \frac{(n+1) !}{n !} =\frac{n+1}{100} \,. \] Choose \(N=101\). Then for all \(n \geq N\), \[\begin{align*} \left|\frac{a_{n+1}}{a_{n}}\right| & = \frac{n+1}{100} \\ & \geq \frac{N+1}{100} \\ & = \frac{101}{100} > 1 \,. \end{align*}\] Hence \(a_{n}\) is divergent by the Ratio Test.

Warning

The Ratio Test in Theorem 159 does not address the case \[ L=1 \,. \] This is because, in this case, the sequence \((a_n)\) might converge or diverge.

For example:

• Define the sequence \[ a_n = \frac1n \,. \] We have \[ \left|\frac{a_{n+1}}{a_{n}}\right| = \frac{n}{n+1} \rightarrow L = 1 \,. \] Hence we cannot apply the Ratio Test. However we know that \[ \lim_{n \to \infty} \, \frac1n = 0 \,. \]

• Consider the sequence \[ a_n = n \,. \] We have \[ \frac{|a_{n+1}|}{|a_n|} = \frac{n+1}{n} \rightarrow L = 1 \,. \] Hence we cannot apply the Ratio Test. However we know that \((a_n)\) is unbounded, and thus divergent.

If the sequence \((a_n)\) is geometric, the Ratio Test of Theorem 159 will give the same answer as the Geometric Sequence Test of Theorem 156. This is the content of the following remark.

Remark 47

Let \(x \in \mathbb{R}\) and define the geometric sequence \[ a_{n}=x^{n} \,. \] Then \[ \left|\frac{a_{n+1}}{a_{n}}\right| = \frac{\left|x^{n+1}\right|}{\left|x^{n}\right|} = \frac{|x|^{n+1}}{|x|^{n}} =|x| \rightarrow |x| \,. \] Hence:

• If \(|x|<1\), the sequence \((a_n)\) converges by the Ratio Test
• If \(|x|>1\), the sequence \((a_n)\) diverges by the Ratio Test.
• If \(|x|=1\), the sequence \((a_n)\) might be convergent or divergent.

These results are in agreement with the Geometric Sequence Test.

6.9 Monotone sequences

We have seen in Theorem 136 that convergent sequences are bounded. We noted that the converse statement is not true. For example the sequence \[ a_n = (-1)^n \] is bounded but not convergent, as shown in Theorem 133. On the other hand, if a bounded sequence is monotone, then it is convergent.

Definition 48: Monotone sequence

Let \((a_n)\) be a real sequence. We say that:

1. \((a_n)\) is increasing if \[ a_n \leq a_{n+1} \,, \quad \forall \, n \geq N \,. \]

2. \((a_n)\) is decreasing if \[ a_n \geq a_{n+1} \,, \quad \forall \, n \geq N \,. \]

3. \((a_n)\) is monotone if it is either increasing or decreasing.

Example 49

Question. Prove that the following sequence is increasing \[ a_n = \frac{n-1}{n} \,. \]

Solution. We have \[ a_{n+1} = \frac{n}{n+1} > \frac{n-1}{n} = a_n \,, \] where the inequality holds because \[\begin{align*} \frac{n}{n+1} > \frac{n-1}{n} \quad & \iff \quad n^2 > (n-1)(n+1) \quad \\ & \iff \quad n^2 > n^2 - 1 \\ & \iff \quad 0 > - 1 \end{align*}\]

Example 50

Question. Prove that the following sequence is decreasing \[ a_n = \frac1n \,. \]

Solution. We have \[ a_n = \frac1n > \frac{1}{n+1} = a_{n+1} \,, \] concluding.

The main result about monotone sequences is the Monotone Convergence Theorem.

Theorem 51: Monotone Convergence Theorem

Let \((a_n)\) be a sequence in \(\mathbb{R}\). Suppose that \((a_n)\) is bounded and monotone. Then \((a_n)\) converges.

Proof

Assume \((a_n)\) is bounded and monotone. Since \((a_n)\) is bounded, the set \[ A:=\{ a_n \, \colon \,n \in \mathbb{N}\} \subseteq \mathbb{R} \] is bounded below and above. By the Axiom of Completeness of \(\mathbb{R}\) there exist \(i,s \in \mathbb{R}\) such that \[ i = \inf A \,, \quad s = \sup A \,. \]

We have two cases:

1. \((a_n)\) is increasing: We are going to prove that \[ \lim_{n \to \infty} a_n = s \,. \] Equivalently, we need to prove that \[ \forall \, \varepsilon>0 \,, \, \exists \, N \in \mathbb{N}\, \text{ s.t. } \, \forall \, n \geq N \,, \,\, |a_n - s| < \varepsilon\,. \tag{6.15}\] Let \(\varepsilon> 0\). Since \(s\) is the smallest upper bound for \(A\), this means \[ s - \varepsilon \] is not an upper bound. Therefore there exists \(N \in \mathbb{N}\) such that \[ s - \varepsilon< a_N \,. \tag{6.16}\] Let \(n \geq N\). Since \(a_n\) is increasing, we have \[ a_N \leq a_n \,, \quad \forall \, n \geq N \,. \tag{6.17}\] Moreover \(s\) is the supremum of \(A\), so that \[ a_n \leq s < s + \varepsilon\,, \quad \forall \, n \in \mathbb{N}\,. \tag{6.18}\] Putting together estimates (A.40)-(A.41)-(A.42) we get \[ s - \varepsilon< a_N \leq a_n \leq s < s + \varepsilon\,, \quad \forall \, n \geq N \,. \] The above implies \[ s - \varepsilon< a_n < s + \varepsilon\,, \quad \forall \, n \geq N \,, \] which is equivalent to (A.39).

2. \((a_n)\) is decreasing: With a similar proof, one can show that \[ \lim_{n \to \infty} a_n = i \,. \] This is left as an exercise.

6.9.1 Example: Euler’s Number

As an application of the Monotone Convergence Theorem we can give a formal definition for the Euler’s Number \[ e = 2.71828182845904523536 \dots \]

Theorem 52

Consider the sequence \[ a_n = \left( 1 + \frac{1}{n} \right)^n \,. \]

We have that:

1. \((a_n)\) is monotone increasing,
2. \((a_n)\) is bounded.

In particular \((a_n)\) is convergent.

Proof

Part 1. We prove that \((a_n)\) is increasing \[ a_{n} \geq a_{n-1} \,, \quad \forall \, n \in \mathbb{N}\,, \] which by definition is equivalent to \[ \left( 1 + \frac{1}{n} \right)^n \geq \left( 1 + \frac{1}{n - 1} \right)^{n-1} \,, \quad \forall \, n \in \mathbb{N}\,. \] Summing the fractions we get \[ \left( \frac{n+1}{n} \right)^n \geq \left( \frac{n}{n-1} \right)^{n-1} \,. \] Multiplying by \(((n-1)/n)^n\) we obtain \[ \left( \frac{n-1}{n} \right)^n \left( \frac{n+1}{n} \right)^n \geq \frac{n-1}{n} \,, \] which simplifies to \[ \left( 1 - \frac{1}{n^2} \right)^n \geq 1 - \frac1n \,, \quad \forall \, n \in \mathbb{N}\,. \tag{6.19}\] Therefore \((a_n)\) is increasing if and only if (A.43) holds. Recall Bernoulli’s inequality from Lemma 157: For \(x \in \mathbb{R}\), \(x>-1\), it holds \[ (1 + x )^n \geq 1 + nx \,, \quad \forall \, n \in \mathbb{N}\,. \] Appliying Bernoulli’s inequality with \[ x = -\frac{1}{n^2} \] yields \[ \left( 1 - \frac{1}{n^2} \right)^n \geq 1 + n \left( -\frac{1}{n^2} \right) = 1 - \frac{1}{n} \,, \] which is exactly (A.43). Then \((a_n)\) is increasing.

Part 2. We have to prove that \((a_n)\) is bounded, that is, that there exists \(M > 0\) such that \[ |a_n| \leq M \,, \quad \forall \, n \in \mathbb{N}\,. \] To this end, introduce the sequence \((b_n)\) by setting \[ b_n := \left( 1 + \frac1n \right)^{n+1} \,. \] The sequence \((b_n)\) is decreasing.

To prove \((b_n)\) is decreasing, we need to show that \[ b_{n-1} \geq b_n \,, \quad \forall \, n \in \mathbb{N}\,. \] By definition of \(b_n\), the above reads \[ \left( 1 + \frac{1}{n-1} \right)^{n} \geq \left( 1 + \frac{1}{n} \right)^{n+1} \,, \quad \forall \, n \in \mathbb{N}\,. \] Summing the terms inside the brackets, the above is equivalent to \[ \left( \frac{n}{n-1} \right)^{n} \geq \left( \frac{n+1}{n} \right)^{n} \left( \frac{n+1}{n} \right) \,. \] Multiplying by \((n/(n+1))^n\) we get \[ \left( \frac{n^2}{n^2 - 1} \right)^{n} \geq \left( \frac{n+1}{n} \right) \,. \] The above is equivalent to \[ \left( 1 + \frac{1}{n^2 - 1} \right)^{n} \geq \left( 1 + \frac{1}{n} \right) \,. \tag{6.20}\] Therefore \((b_n)\) is decreasing if and only if (A.44) holds for all \(n \in \mathbb{N}\). By choosing \[ x = \frac{1}{n^2 - 1} \] in Bernoulli’s inequality, we obtain \[\begin{align*} \left( 1 + \frac{1}{n^2 - 1} \right)^{n} & \geq 1 + n \left( \frac{1}{n^2 - 1} \right) \\ & = 1 + \frac{n}{n^2 - 1} \\ & \geq 1 + \frac1n \,, \end{align*}\] where in the last inequality we used that \[ \frac{n}{n^2 - 1} > \frac1n \,, \] which holds, being equivalent to \(n^2 > n^2 - 1\). We have therefore proven (A.44), and hence \((b_n)\) is decreasing.

We now observe that For all \(n \in \mathbb{N}\) \[\begin{align*} b_n & = \left( 1 + \frac1n \right)^{n+1} \\ & = \left( 1 + \frac1n \right)^n \left( 1 + \frac1n \right) \\ & = a_n \left( 1 + \frac1n \right) \\ & > a_n \,. \end{align*}\] Since \((a_n)\) is increasing and \((b_n)\) is decreasing, in particular \[ a_n \geq a_1 \,, \quad b_n \leq b_1 \,. \] Therefore \[ a_1 \leq a_n < b_n \leq b_1 \,, \quad \forall \, n \in \mathbb{N}\,. \] We compute \[ a_1 = 2 \,, \quad b_1 = 4 \,, \] from which we get \[ 2 \leq a_n \leq 4 \,, \quad \forall \, n \in \mathbb{N}\,. \] Therefore \[ |a_n| \leq 4 \,, \quad \forall \, n \in \mathbb{N}\,, \] showing that \((a_n)\) is bounded.

Part 3. The sequence \((a_n)\) is increasing and bounded above. Therefore \((a_n)\) is convergent by the Monotone Convergence Theorem 167.

Thanks to Theorem 168 we can define the Euler’s Number \(e\).

Definition 53: Euler’s Number

The Euler’s number is defined as \[ e := \lim_{n \to \infty } \, \left( 1 + \frac{1}{n} \right)^n \,. \]

Setting \(n=1000\) in the formula for \((a_n)\), we get an approximation of \(e\): \[ e \approx a_{1000} = 2.7169 \,. \]

6.10 Special limits

In this section we investigate limits of some sequences to which the Limit Tests do not apply.

Theorem 54

Let \(x \in \mathbb{R}\), with \(x > 0\). Then \[ \lim_{n \to \infty} \sqrt[n]{x} = 1 \,. \]

Proof

Step 1. Assume \(x \geq 1\). In this case \[ \sqrt[n]{x} \geq 1 \,. \] Define \[ b_n := \sqrt[n]{x} - 1 \,, \] so that \(b_n \geq 0\). By Bernoulli’s Inequality we have \[ x = (1 + b_n)^n \geq 1 + n b_n \,. \] Therefore \[ 0 \leq b_n \leq \frac{x-1}{n} \,. \] Since \[ \frac{x-1}{n} \longrightarrow 0 \,, \] by the Squeeze Theorem we infer \(b_n \to 0\), and hence \[ \sqrt[n]{x} = 1 + b_n \longrightarrow 1 + 0 = 1\,, \] by the Algebra of Limits.

Step 2. Assume \(0< x < 1\). In this case \[ \frac{1}{x} > 1 \,. \] Therefore \[ \lim_{n \to \infty} \, \sqrt[n]{ 1/x } = 1 \,. \] by Step 1. Therefore \[ \sqrt[n]{x} = \frac{1}{\sqrt[n]{ 1/x }} \longrightarrow \frac{1}{1} = 1\,, \] by the Algebra of Limits.

Theorem 55

Let \((a_n)\) be a sequence such that \(a_n \to 0\). Then \[ \sin(a_n) \to 0 \,, \quad \cos(a_n) \to 1 \,. \]

Proof

Assume that \(a_n \to 0\) and set \[ \varepsilon:= \frac{\pi}{2} \,. \] By the convergence \(a_m \to 0\) there exists \(N \in \mathbb{N}\) such that \[ |a_n| < \varepsilon= \frac{\pi}{2} \, \quad \forall \, n \geq N \,. \tag{6.21}\]

Step 1. We prove that \[ \sin(a_n) \to 0 \,. \] By elementary trigonometry we have \[ 0 \leq |\sin (x)| = \sin |x| \leq |x| \,, \quad \forall \, x \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \,. \] Therefore, since (A.45) holds, we can substitute \(x=a_n\) in the above inequality to get \[ 0 \leq |\sin (a_n)| \leq |a_n| \,, \quad \forall \, n \geq \mathbb{N}\,. \] Since \(a_n \to 0\), we also have \(|a_n|\to 0\). Therefore \(|\sin (a_n)| \to 0\) by the Squeeze Theorem. This immediately implies \(\sin(a_n) \to 0\).

Step 2. We prove that \[ \cos(a_n) \to 1 \,. \] Inverting the relation \[ \cos^2(x) + \sin^2 (x) = 1 \,, \] we obtain \[ \cos(x) = \pm \sqrt{ 1 - \sin^2 (x) } \,. \] We have that \(\cos(x) \geq 0\) for \(-\pi/2 \leq x \leq \pi/2\). Thus \[ \cos(x) = \sqrt{ 1 - \sin^2 (x) } \,, \quad \forall \, x \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \,. \] Since (A.45) holds, we can set \(x=a_n\) in the above inequality and obtain \[ \cos(a_n) = \sqrt{ 1 - \sin^2 (a_n) } \,, \quad \forall \, n \geq N \,. \] By Step 1 we know that \(\sin(a_n) \to 0\). Therefore, by the Algebra of Limits, \[ 1 - \sin^2 (a_n) \longrightarrow 1 - 0 \cdot 0 = 1 \,. \] Using Theorem 148 we have \[ \cos(a_n) = \sqrt{1 - \sin^2 (a_n)} \longrightarrow \sqrt{1} = 1 \,, \] concluding the proof.

Theorem 56

Suppose \((a_n)\) is such that \(a_n \to 0\) and \[ a_n \neq 0 \,, \quad \forall \, n \in \mathbb{N}\, . \] Then \[ \lim_{n \to \infty} \frac{\sin(a_n)}{a_n} = 1 \,. \]

Proof

The following elementary trigonometric inequality holds: \[ \sin (x) < x < \tan (x) \,, \quad \forall \, x \in \left[0 ,\frac{\pi}{2} \right] \,. \] Note that \(\sin x >0\) for \(0 < x < \pi/2\). Therefore we can divide the above inequality by \(\sin(x)\) and take the reciprocals to get \[ \cos(x) < \frac{\sin (x)}{x} < 1 \,, \quad \forall \, x \in \Big( 0 ,\frac{\pi}{2} \Big] \,. \] If \(-\pi/2<x<0\), we can apply the above inequality to \(-x\) to obtain \[ \cos(-x) < \frac{\sin(-x)}{-x} < 1 \,. \] Recalling that \(\cos(-x) = \cos(x)\) and \(\sin(-x)=-\sin(x)\), we get \[ \cos(x) < \frac{\sin(x)}{x} < 1 \,, \quad \forall \, x \in \Big( -\frac{\pi}{2}, 0 \Big] \,. \] Thus \[ \cos(x) < \frac{\sin(x)}{x} < 1 \,, \quad \forall \, x \in \left[-\frac{\pi}{2}, \frac{\pi}{2} \right] \smallsetminus \{0\} \,. \tag{6.22}\] Let \[ \varepsilon:= \frac{\pi}{2} \,. \] Since \(a_n \to 0\), there exists \(N \in \mathbb{N}\) such that \[ |a_n| < \varepsilon= \frac{\pi}{2} \,, \quad \forall \, n \geq N \,. \] Since \(a_n \neq 0\) by assumption, the above shows that \[ a_n \in \Big[ -\frac{\pi}{2}, \frac{\pi}{2} \Big] \smallsetminus \{0\} \,, \quad \forall \, n \geq \mathbb{N}\,. \] Therefore we can substitute \(x=a_n\) into (A.46) to get \[ \cos(a_n) < \frac{\sin(a_n)}{a_n} < 1 \,, \quad \forall \, n \geq N \,. \] We have \[ \cos(a_n) \to 1 \] by Theorem 171. By the Squeeze Theorem we conclude that \[ \lim_{n \to \infty} \frac{\sin(a_n)}{a_n} = 1 \,. \]

Warning

You might be tempted to apply L’Hôpital’s rule (which we did not cover in these Lecture Notes) to compute \[ \lim_{x \to 0} \frac{\sin(x)}{x} \,. \] This would yield the correct limit \[ \lim_{x \to 0} \frac{\sin(x)}{x} = \lim_{x \to 0} \frac{(\sin(x))'}{(x)'} = \lim_{x \to 0} \cos(x) = 1 \,. \] However this is a circular argument, since the derivative of \(\sin(x)\) at \(x=0\) is defined as the limit \[ \lim_{x \to 0} \frac{\sin(x)}{x} \,. \]

Theorem 57

Suppose \((a_n)\) is such that \(a_n \to 0\) and \[ a_n \neq 0 \,, \quad \forall \, n \in \mathbb{N}\, . \] Then \[ \lim_{n \to \infty} \frac{1 - \cos(a_n)}{(a_n)^2} = \frac{1}{2} \,, \quad \lim_{n \to \infty} \frac{1 - \cos(a_n)}{a_n} = 0 \, . \]

Proof

Step 1. By Theorem 171 and Theorem 172, we have \[ \cos(a_n) \to 1 \,, \quad \frac{\sin(a_n)}{a_n} \to 1 \,. \] Therefore \[\begin{align*} \frac{1 - \cos(a_n)}{(a_n)^2} & = \frac{1 - \cos(a_n)}{(a_n)^2} \, \frac{1 + \cos(a_n)}{1 + \cos(a_n)} \\ & = \frac{1 - \cos^2(a_n)}{(a_n)^2} \, \frac{1}{1 + \cos(a_n)} \\ & = \left( \frac{\sin(a_n)}{a_n} \right)^2 \, \frac{1}{1 + \cos(a_n)} \longrightarrow 1 \cdot \frac{1}{1 + 1} = \frac12 \,, \end{align*}\] where in the last line we use the Algebra of Limits.

Step 2. We have \[ \frac{1 - \cos(a_n)}{a_n} = a_n \cdot \frac{1 - \cos(a_n)}{(a_n)^2} \longrightarrow 0 \cdot \frac{1}{2} = 0 \,, \] using Step 1 and the Algebra of Limits.

Example 58

Question. Prove that \[ \lim_{n \to \infty} \, n \sin \left( \frac{1}{n} \right) = 1 \,. \tag{6.23}\]

Solution. This is because \[ n \sin \left( \frac{1}{n} \right) = \frac{ \sin \left( \dfrac{1}{n} \right) }{ \dfrac{1}{n} } \longrightarrow 1 \,, \] by Theorem 172 with \(a_n = 1/n\).

Example 59

Question. Prove that \[ \lim_{n \to \infty} \, n^2 \left( 1 - \cos \left( \dfrac{1}{n} \right) \right) = \frac12 \,. \tag{6.24}\]

Solution. Indeed, \[ n^2 \left( 1 - \cos \left( \frac{1}{n} \right)\right) = \dfrac{1 - \cos \left( \dfrac{1}{n} \right)}{\dfrac{1}{n^2}} \longrightarrow \frac12 \,, \] by applying Theorem 173 with \(a_n = 1/n\).

Example 60

Question. Prove that \[ \lim_{n \to \infty} \, \frac{n \left( 1- \cos \left( \dfrac{1}{n} \right) \right) }{ \sin \left( \dfrac{1}{n} \right) } = \frac12 \,. \]

Solution. Using (A.48)-(A.47) and the Algebra of Limits \[\begin{align*} \frac{n \left( 1- \cos \left( \dfrac{1}{n} \right) \right) }{ \sin \left( \dfrac{1}{n} \right) } & = \frac{n^2 \left( 1- \cos \left( \dfrac{1}{n} \right) \right) }{ n \sin \left( \dfrac{1}{n} \right) } \\ & \longrightarrow \frac{1/2}{1} = \frac12 \,. \end{align*}\]

Example 61

Question. Prove that \[ \lim_{n \to \infty} \, n \cos \left( \frac{2}{n} \right) \sin \left( \frac{2}{n} \right) = 2 \,. \]

Solution. We have \[ \cos \left( \frac{2}{n} \right) \longrightarrow 1 \,, \] by Theorem 171 applied with \(a_n = 2/n\). Moreover \[ \frac{\sin \left( \dfrac{2}{n} \right)}{\dfrac{2}{n}} \longrightarrow 1 \,, \] by Theorem 172 applied with \(a_n = 2/n\). Therefore \[\begin{align*} n \cos \left( \frac{2}{n} \right) \sin \left( \frac{2}{n} \right) & = 2 \cdot \cos \left( \frac{2}{n} \right) \cdot \frac{\sin \left( \dfrac{2}{n} \right)}{\dfrac{2}{n}} \\ & \longrightarrow 2 \cdot 1 \cdot 1 = 2 \,, \end{align*}\] where we used the Algebra of Limits.

Example 62

Question. Prove that \[ \lim_{n \to \infty} \, \frac{n^2+1}{n+1} \sin \left( \dfrac{1}{n} \right) = 1 \,. \]

Solution. Note that \[\begin{align*} \frac{n^2+1}{n+1} \sin \left( \dfrac{1}{n} \right) & = \left( \frac{1+\dfrac{1}{n^2}}{1+ \dfrac{1}{n}} \right) \cdot \left( n \sin \left( \dfrac{1}{n} \right) \right) \\ & \longrightarrow \frac{1 + 0}{1 + 0} \cdot 1 = 1\,, \end{align*}\] where we used (A.47) and the Algebra of Limits.