5 Sequences in \(\mathbb{R}\)

Definition 1: 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.

Theorem 2

Let \(p>0\). Then \[ \lim _{n \rightarrow \infty} \frac{1}{n^{p}}=0 \,. \]

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 \[ N > \frac{1}{\varepsilon^{1 / p}} \,. \tag{5.1}\] 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 (5.1) we deduce \[ \frac{1}{N^p} < \varepsilon\,. \] Then \[ \left|\frac{1}{n^{p}}-0\right|=\frac{1}{n^{p}} \leq \frac{1}{N^{p}} < \varepsilon\,. \]

Example 3

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

Solution.

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 \[ N > \frac{3}{4\varepsilon} - \frac{6}{4} \,. \]
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 \[ N > \frac{3}{4\varepsilon} - \frac{6}{4} \,. \tag{5.2}\] By the rough work shown above, inequality (5.2) 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 \(n \geq N\).

Definition 4: 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.

Theorem 5

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

Proof

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.

Theorem 6: 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 \(a=b\).

Definition 7: 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 \[ \left|a_{n}\right| \leq M \,, \quad \forall \, n \in \mathbb{N}\,. \]

Theorem 8

Every convergent sequence is bounded.

Example 9

The sequence \[ a_n = (-1)^n \] is bounded but not convergent.

Corollary 10

If a sequence is not bounded, then the sequence does not converge.

Remark 11

For a sequence \((a_n)\) to be unbounded, it means that \[ \forall \, M > 0 \,, \,\, \exists \, n \in \mathbb{N}\, \text{ s.t. } \, \left|a_{n}\right| > M \,. \]

Theorem 12

For all \(p>0\), the sequence \[ a_n = n^p \] does not converge.

Theorem 13

The sequence \[ a_n = \log n \] does not converge.

Theorem 14: 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,

Limit of sum is the sum of limits: \[ \lim_{n \rightarrow \infty}\left(a_{n} \pm b_{n}\right)=a \pm b \]
Limit of product is the product of limits: \[ \lim _{n \rightarrow \infty}\left(a_{n} b_{n}\right) = a b \]
If \(b_{n} \neq 0\) for all \(n \in \mathbb{N}\) and \(b \neq 0\), then \[ \lim_{n \rightarrow \infty} \left(\frac{a_{n}}{b_{n}}\right)=\frac{a}{b} \]

Example 15

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

Solution. We can rewrite \[ \frac{3 n}{7 n+4}=\frac{3}{7+\dfrac{4}{n}} \] From Theorem 2, we know that \[ \frac{1}{n} \rightarrow 0 \,. \] Hence, it follows from Theorem 14 Point 2 that \[ \frac{4}{n} = 4 \cdot \frac1n \rightarrow 4 \cdot 0 = 0 \,. \] By Theorem 14 Point 1 we have \[ 7 + \frac{4}{n} \rightarrow 7 + 0 = 7 \,. \] Finally we can use Theorem 14 Point 3 to infer \[ \frac{3 n}{7 n+4}=\frac{3}{7+\dfrac{4}{n}} \rightarrow \frac{3}{7} \,. \]

Example 16

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 2 we have \[ \frac{1}{n^2} \to 0 \,. \] We can then use the Algebra of Limits Theorem 14 Point 2 to infer \[ \frac{3}{n^2} \to 3 \cdot 0 = 0 \] and Theorem 14 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 14 Point 3 and conclude \[ \frac{1-\dfrac{1}{n^{2}}}{2-\dfrac{3}{n^{2}}} \to \frac{1}{2} \,. \] Therefore \[ \lim_{n \to \infty } \, \frac{n^{2}-1}{2 n^{2}-3} = \lim_{n \to \infty} \, \frac{1-\dfrac{1}{n^{2}}}{2-\dfrac{3}{n^{2}}} = \frac{1}{2} \,. \]

Example 17

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 14 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, 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 14 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.

Example 18

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 14 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 14 Point 2 ensures that \[ a_{n} \to \frac{8}{5} \cdot \frac{1}{3}=\frac{8}{15} \,. \]

Example 19

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.

Theorem 20

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 \[ \lim _{n \rightarrow \infty} \sqrt{a_{n}}=\sqrt{a} \,. \]

Example 21

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 20 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 22

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 20. 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.

5.1 Limit Tests

Theorem 23: 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 \[ \lim_{n \rightarrow \infty} a_{n}= L \, . \]

Example 24

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 23 we conclude \[ \lim_{n \to \infty} \frac{(-1)^{n}}{n}=0 \,. \]

Example 25

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: \[ -1 + 9 n^{2} \leq \cos (3 n)+9 n^{2} \leq 1+ 9 n^{2} \,. \tag{5.3}\] Concerning the denominator, we have \[ 11 n^{2}-15 \leq 11 n^{2}+15 \sin (17 n) \leq 11 n^{2} + 15 \] and therefore \[ \frac{1}{11 n^{2} + 15} \leq \frac{1}{11 n^{2}+15 \sin (17 n)} \leq \frac{1}{11 n^{2}-15} \,. \tag{5.4}\] Putting together (5.3)-(5.4) 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 23 we conclude \[ \lim_{n \to \infty} \frac{\cos (3 n)+ 9 n^{2}}{11 n^{2}+15 \sin (17 n)}=\frac{9}{11} \,. \]

Theorem 26: Geometric Sequence Test

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

If \(|x|<1\), then \[ \lim_{n \to \infty} a_{n} = 0 \,. \]
If \(|x|>1\), then sequence \(\left(a_{n}\right)\) is unbounded, and hence divergent.

Example 27

We can apply Theorem 26 to prove convergence or divergence for the following sequences.

We have \[ \left(\frac{1}{2}\right)^{n} \longrightarrow 0 \] since \[ \left|\frac{1}{2}\right|=\frac{1}{2}<1 \,. \]
We have \[ \left(\frac{-1}{2}\right)^{n} \longrightarrow 0 \] since \[ \left|\frac{-1}{2}\right|=\frac{1}{2}<1 \,. \]
The sequence \[ a_n = \left(\frac{-3}{2}\right)^{n} \] does not converge, since \[ \left|\frac{-3}{2}\right|=\frac{3}{2}>1 \,. \]
As \(n \rightarrow \infty\), \[ \frac{3^{n}}{(-5)^{n}}=\left(-\frac{3}{5}\right)^{n} \longrightarrow 0 \] since \[ \left|-\frac{3}{5}\right|=\frac{3}{5}<1 \,. \]
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 \[ \left|-\frac{7}{4}\right|=\frac{7}{4}>1 \,. \]

Theorem 28: Ratio Test

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

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.
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.

Example 29

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 28.

Example 30

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 31

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.

5.2 Monotone sequences

Definition 32: Monotone sequence

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

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

Example 33

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 34

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.

Theorem 35: 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:

\((a_n)\) is increasing: We are going to prove that \[ \lim_{n \to \infty} a_n = s \,. \] Equivalently, we need to prove that \[ \forall \, \varepsilon>0 \,, \, \exists \, N \in \mathbb{N}\, \text{ s.t. } \, \forall \, n \geq N \,, \,\, |a_n - s| < \varepsilon\,. \tag{5.5}\] 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 \[ s - \varepsilon< a_N \,. \tag{5.6}\] Let \(n \geq N\). Since \(a_n\) is increasing, we have \[ a_N \leq a_n \,, \quad \forall \, n \geq N \,. \tag{5.7}\] Moreover \(s\) is the supremum of \(A\), so that \[ a_n \leq s < s + \varepsilon\,, \quad \forall \, n \in \mathbb{N}\,. \tag{5.8}\] Putting together estimates (5.6)-(5.7)-(5.8) 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 (5.5).
\((a_n)\) is decreasing: With a similar proof, one can show that \[ \lim_{n \to \infty} a_n = i \,. \] This is left as an exercise.

5.3 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 36

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

We have that:

\((a_n)\) is monotone increasing,
\((a_n)\) 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 \[ \left( 1 - \frac{1}{n^2} \right)^n \geq 1 - \frac1n \,, \quad \forall \, n \in \mathbb{N}\,. \tag{5.9}\] Therefore \((a_n)\) is increasing if and only if (5.9) holds. Recall Bernoulli’s inequality from Lemma \(\ref{lemma-bernoulli}\): 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 (5.9). 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 \[ \left( 1 + \frac{1}{n^2 - 1} \right)^{n} \geq \left( 1 + \frac{1}{n} \right) \,. \tag{5.10}\] Therefore \((b_n)\) is decreasing if and only if (5.10) 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 (5.10), 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 35.

Thanks to Theorem 36 we can define the Euler’s Number \(e\).

Definition 37: 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 \(e\): \[ e \approx a_{1000} = 2.7169 \,. \]

5.4 Some important limits

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

Theorem 38

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 39

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 \[ |a_n| < \varepsilon= \frac{\pi}{2} \, \quad \forall \, n \geq N \,. \tag{5.11}\]

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 (5.11) 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 (5.11) 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 20 we have \[ \cos(a_n) = \sqrt{1 - \sin^2 (a_n)} \longrightarrow \sqrt{1} = 1 \,, \] concluding the proof.

Theorem 40

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 \[ \cos(x) < \frac{\sin(x)}{x} < 1 \,, \quad \forall \, x \in \left[-\frac{\pi}{2}, \frac{\pi}{2} \right] \smallsetminus \{0\} \,. \tag{5.12}\] 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 (5.12) 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 39. 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 41

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 39 and Theorem 40, 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 42

Question. Prove that \[ \lim_{n \to \infty} \, n \sin \left( \frac{1}{n} \right) = 1 \,. \tag{5.13}\]

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 40 with \(a_n = 1/n\).

Example 43

Question. Prove that \[ \lim_{n \to \infty} \, n^2 \left( 1 - \cos \left( \dfrac{1}{n} \right) \right) = \frac12 \,. \tag{5.14}\]

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 41 with \(a_n = 1/n\).

Example 44

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 (5.14)-(5.13) 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 45

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 39 applied with \(a_n = 2/n\). Moreover \[ \frac{\sin \left( \dfrac{2}{n} \right)}{\dfrac{2}{n}} \longrightarrow 1 \,, \] by Theorem 40 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 46

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 (5.13) and the Algebra of Limits.