8 Series

A series is the sum of all terms in a sequence: \[ a_1 + a_2 + a_3 + \ldots + a_n + \ldots \] Since we are dealing with an infinite amount of terms, we need to be careful. For example, consider the series \[ \sum_{n=1}^\infty (-1)^n = -1 + 1 - 1 + 1 - 1 + 1 - \ldots \tag{8.1}\] If we sum the terms in pairs, we obtain \[ \sum_{n=1}^\infty (-1)^n = (-1 + 1) + (- 1 + 1) + ( - 1 + 1) + \ldots = 0\,. \] If we reorder the terms, we obtain a different result \[\begin{align*} \sum_{k=1}^\infty (-1)^n & = (1 + 1) - 1 + (1 + 1) - 1 + \ldots \\ & = (2 - 1) + (2 - 1) + \ldots \\ & = 1 + 1 + \ldots \\ & = \infty \,. \end{align*}\] Therefore commutativity of the sum does not hold when summing infinitely many terms. We need a good definition of convergence.

8.1 Convergent series

We will develop the theory of series for sequences in \(\mathbb{C}\). All the results for complex series will also hold for series in \(\mathbb{R}\). Some results will only hold for series in \(\mathbb{R}\): this will be the case for comparison tests in which the order relation of \(\mathbb{R}\) is needed.

We start by defining partial sums.

Definition 1: Partial sums

Let \((a_{n})\) be a sequence in \(\mathbb{C}\). The \(k\)-th partial sum of \((a_n)\) is \[ s_{k} :=a_{1}+a_{2}+\ldots+a_{k} = \sum_{n=1}^{k} a_{n} \] This sequence \(\left(s_{k}\right)_{k \in \mathbb{N}}\) is called the sequence of partial sums.

We can use the sequence of partial sums to define convergence of a series.

Definition 2: Convergent series

Let \((a_{n})\) be a sequence in \(\mathbb{C}\). We denote the series of \(\left(a_{n}\right)_{n \in \mathbb{N}}\) by \[ \sum_{n=1}^{\infty} a_{n} \] We say that this series converges to \(s \in \mathbb{C}\) if \[ \lim_{k \rightarrow \infty} \sum_{n=1}^{k} a_{n} = \lim_{k \rightarrow \infty} s_{k} = s \,. \] In this case we write \[ \sum_{n=1}^{\infty} a_{n} = s \,. \]

Definition 3: Divergent series

Let \((a_{n})\) be a sequence in \(\mathbb{C}\). The series \[ \sum_{n=1}^\infty a_n \] is divergent if the sequence of partial sums \((s_k)\) is divergent.

Example 4

Consider the series \[ \sum_{n=1}^{\infty} \frac{1}{n(n+1)} \,. \] Note that \[ \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1} \,. \] Hence, we can compute the partial sums \(s_{n}\) as follows: \[\begin{align*} s_{k} & =\sum_{n=1}^{k} \frac{1}{n(n+1)} \\ & =\sum_{n=1}^{k}\left(\frac{1}{n}-\frac{1}{n+1}\right) \\ & =\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{k}-\frac{1}{k+1} \\ & =1-\frac{1}{k+1} . \end{align*}\] Since, \[ \lim_{k \rightarrow \infty} s_{k}=\lim _{k \rightarrow \infty}\left(1-\frac{1}{k+1}\right)=1 \,, \] the series converges to \(1\), that is, \[ \sum_{n=1}^{\infty} \frac{1}{n(n+1)}=1 \,. \] A series of this kind is called a telescopic sum, since we can fold the entire partial sum together, in such a way that only two terms remain.

Let us go back to the series considered in (8.1).

Example 5

Consider the series \[ \sum_{n=1}^{\infty}(-1)^{n} \,. \] The partial sums \(s_{k}\) are given by \[ s_{k}=\sum_{n=1}^{k}(-1)^{n}= \begin{cases}-1 & \text { if } \, n \, \text { is odd } \, \\ 0 & \text { if } \, n \, \text { is even. }\end{cases} \] Therefore \(s_{k}\) diverges, so also the series diverges.

In general, it is a difficult taks to compute the exact value of a series. Therefore, we will mainly focus our effort on determining whether a series converges or not.

The following theorem shows that if the terms in the sequence do not converge to \(0\), then the series cannot converge.

Theorem 6: Necessary Condition for Convergence

Let \((a_{n})\) be a sequence in \(\mathbb{C}\). If the series \[ \sum_{n=1}^{\infty} a_{n} \] converges, then \[ \lim_{n \rightarrow \infty} a_{n}=0 \,. \]

Proof

Suppose that \[ \sum_{n=1}^{\infty} a_{n} \] converges. By definition of convergent series there exists some \(s \in \mathbb{C}\) such that \[ \lim_{k \rightarrow \infty} s_{k}=\lim_{k \rightarrow \infty} \sum_{n=1}^{k} a_{n} = s \,. \] Then also \[ \lim_{k \rightarrow \infty} s_{k+1} = s \,. \] Hence, by the Algebra of Limits in \(\mathbb{C}\), we have that \[ \lim_{k \rightarrow \infty}\left(s_{k+1}-s_{k}\right)=\lim _{k \rightarrow \infty} s_{k+1}-\lim_{k \rightarrow \infty} s_{k} = s - s = 0 \,. \] Noting that \[ s_{k+1} - s_{k} = a_{k+1}\,, \quad \forall \, k \in \mathbb{N}\,, \] we obtain \[ \lim_{k \rightarrow \infty} a_{k} = \lim_{k \rightarrow \infty} a_{k+1} = \lim_{k \rightarrow \infty} \left( s_{k+1} - s_{k} \right) = 0 \, . \]

Important

Theorem 6 is saying that, if \[ \lim_{n \rightarrow \infty} a_{n} \neq 0 \,, \] then the series \[ \sum_{n=1}^{\infty} a_{n} \] does not converge.

Example 7

Consider the series \[ \sum_{n=1}^\infty (-1)^n \,. \tag{8.2}\] We have that \[ \lim_{n \to \infty} a_n = \lim_{n \to \infty} (-1)^n \neq 0 \,, \] being \((a_n)\) divergent. Therefore the series at (8.2) diverges by Theorem 6.

Example 8

Consider the series \[ \sum_{n=1}^{\infty} \frac{n}{5 n+11} \,. \] Then \[ a_{n}:=\frac{n}{5 n+11}=\frac{1}{5 + \dfrac{11}{n} } \longrightarrow \frac{1}{5} \neq 0 \,. \] Hence, the series does not converge.

Important

Theorem 6 says that if \(\sum_{n=1}^\infty a_n\) converges, then \[ a_n \to 0 \,. \] The converse is false: In general the condition \(a_n \to 0\) does not guarantee convergence of the associated series, as shown in the example below.

Example 9

Consider the series \[ \sum_{n=1}^\infty a_n \,, \quad a_n := \frac{1}{ \sqrt{n+1} + \sqrt{n} } \,. \] By the Algebra of Limits we have \[ \lim_{n \to \infty} a_n = 0 \,. \] However the partial sums are given by \[ s_k = \sum_{n=1}^k a_n = \sqrt{k+1} - 1 \,. \] Therefore \((s_k)\) is divergent and so the series is divergent.

Exercise: Prove by induction that \[ \sum_{n=1}^k a_n = \sqrt{k+1} - 1 \,. \]

Remark 10

It is customary to sum a series starting at \(n=1\). However one could start the sum at any \(n=N\) with \(N \in \mathbb{N}\). This does not affect the convergence of the series, in the sense that \[ \sum_{n=1}^\infty a_n \,\, \mbox{ converges \,} \iff \,\, \sum_{n=N}^\infty a_n \,\, \mbox{ converges.} \] In case of convergence, we would of course have \[ \sum_{n=N}^\infty a_n = \sum_{n=1}^\infty a_n - \left( a_1 + \ldots + a_{N-1} \right) \,. \]

Example 11

We have seen that \[ \sum_{n=1}^{\infty} \frac{1}{n(n+1)} = 1 \,. \] Hence also the series \[ \sum_{n=7}^{\infty} \frac{1}{n(n+1)} \] converges. However, in this case, the partial sums are given by \[\begin{align*} s_{k} & =\sum_{n=7}^{k} \frac{1}{n(n+1)} \\ & =\sum_{n=7}^{k}\left(\frac{1}{n}-\frac{1}{n+1}\right) \\ & =\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\ldots+\frac{1}{k}-\frac{1}{k+1} \\ & =\frac{1}{7}-\frac{1}{k+1} \,. \end{align*}\] Therefore \[ \sum_{n=7}^{\infty} \frac{1}{n(n+1)} = \lim_{k \rightarrow \infty} s_{k} = \frac{1}{7} \,. \]

8.2 Geometric series

Definition 12: Geometric Series in \(\mathbb{C}\)

Let \(x \in \mathbb{C}\). The geometric series of ratio \(x\) is the series \[ \sum_{n=0}^{\infty} x^{n} \,. \]

Geometric series are one of the few types of series that can be explicitly computed, as stated in the following theorem.

Theorem 13: Geometric Series Test

Let \(x \in \mathbb{C}\). We have:

If \(|x|<1\), then the geometric series of ratio \(x\) converges, with

\[ \sum_{n=0}^{\infty} x^{n}=\frac{1}{1-x} \,. \tag{8.3}\]

If \(|x| \geq 1\), then the geometric series of ratio \(x\) diverges.

Important

Recall that the Geometric Sequence Test does not cover the case \(|x|=1\), since in general the sequence \[ a_n = x^n \] could be convergent or divergent. However the Geometric Series Test covers the case \(|x|=1\), in which case \[ \sum_{n=0}^{\infty} x^{n} \] diverges.

Proof

Part 1. Suppose that \(|x|<1\). By using induction we prove that \[ s_{k} := \sum_{n=0}^{k} x^{n}=\frac{1-x^{k+1}}{1-x} \,, \quad \forall \, k \in \mathbb{N}\,. \tag{8.4}\]

Base case: For \(k=0\), we get that \[ s_{0} = x^{0} = 1 = \frac{1-x^{1}}{1-x} \,. \]
Induction step: Let \(k \in \mathbb{N}\cup \{0\}\) and suppose that \[ s_{k}=\frac{1-x^{k+1}}{1-x} \,. \] Then, \[\begin{align*} s_{k+1} & = s_{k}+x^{k+1} \\ & = \frac{1-x^{k+1}}{1-x}+x^{k+1} \\ & = \frac{1-x^{k+1}+(1-x) x^{k+1}}{1-x} \\ & = \frac{1-x^{k+1}+x^{k+1}-x^{k+2}}{1-x} \\ & = \frac{1-x^{k+2}}{1-x} \,, \end{align*}\] concluding the proof of the inductive step.

By the Principle of Induction formula (8.4) holds for all \(k \in \mathbb{N}\). Since \(|x|<1\), by the Geometric Sequence Test we infer \[ \lim_{k \to \infty} \, x^{k} = 0 \,. \] Hence \[\begin{align*} \sum_{n=0}^{\infty} x^{n} & = \lim_{k \to \infty } \, s_k \\ & = \lim_{k \rightarrow \infty} \, \frac{1-x^{k+1}}{1-x} \\ & = \lim_{k \rightarrow \infty} \frac{1-x \cdot x^{k}}{1-x} \\ & = \frac{1}{1-x} \,, \end{align*}\] where the last equality follows from the Algebra of Limits.

Part 2. Suppose \(|x| \geq 1\). Then \[ \lim_{n \to \infty} \, x^n \neq 0 \,. \tag{8.5}\] Indeed, suppose by contradiction that \(x_n \to 0\). Hence, for \(\varepsilon= 1/2\), there exists \(N \in \mathbb{N}\) such that \[ \left|x^{n}-0\right| < \varepsilon= \frac{1}{2} \,, \quad \forall \, n \geq \mathbb{N}\,. \] However \[ \left|x^{n}-0\right| = |x^n| = |x|^n \geq 1 \,, \] which yields \[ 1 < \varepsilon= \frac12 \,, \] contradiction. Then (8.5) holds and the series \[ \sum_{n=0}^{\infty} x^{n} \] diverges by the Necessary Condition in Theorem 6.

Let us show some applications of the Geometric Series Test of Theorem 13.

Example 14

Since \(\left|\frac{1}{2}\right|<1\) we have \[ \sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{n}=\frac{1}{1-\dfrac{1}{2}}=2 \]
Since \(\left|\frac{-3}{2}\right|=\frac{3}{2}>1\) the series \[ \sum_{n=0}^{\infty}\left(\frac{-3}{2}\right)^{n} \] does not converge.
Since \(\left|\frac{-3}{4}\right|=\frac{3}{4}<1\), we have \[ \sum_{n=0}^{\infty}\left(\frac{-3}{4}\right)^{n}=\frac{1}{1-\dfrac{-3}{4}}=\frac{1}{\dfrac{7}{4}}=\frac{4}{7} \]
Since \(|-1|=1\), the series \[ \sum_{n=0}^{\infty}(-1)^{n} \] does not converge.

Remark 15

If the sum of a Geometric Sries does not start at \(n=0\), we need to tweak the summation formula at (8.3). For example, if \(|x|<1\), and we start the series at \(n=1\), we get \[ \sum_{n=1}^\infty x^k = \frac{1}{1-x} - 1 = \frac{x}{1-x} \,. \]

Example 16

We have that

\[\sum_{n=1}^{\infty}\left(\frac{1}{2}\right)^{n}= \frac{ \dfrac{1}{2}}{1- \dfrac{1}{2}} = 1 \,. \]

The Geometric Series Test of Theorem 13 can be applied to complex geometric series as well.

Example 17

Consider the series \[ \sum_{n=0}^{\infty} \frac{1}{(1+i)^{n}} \,. \] Since \[ \frac{1}{(1+i)^{n}}=\left(\frac{1}{1+i}\right)^{n} \] and \[ \left|\frac{1}{1+i}\right| = \frac{1}{\sqrt{1^{2}+1^{2}}} = \frac{1}{\sqrt{2}} < 1\,, \] Therefore the series converges by the Geometric Series Test, and \[\begin{align*} \sum_{n=0}^{\infty} \frac{1}{(1+i)^{n}} & = \frac{1}{1-\dfrac{1}{1+i}} \\ & = \frac{1}{\dfrac{1+i-1}{1+i}} \\ & = \frac{1+i}{i} \\ & = \frac{(1+i) i}{i^{2}} \\ & = \frac{i-1}{-1} \\ & = 1-i \,. \end{align*}\]
Consider the series \[ \sum_{n=0}^{\infty}\left(\frac{-5-5 i}{3+3 i}\right)^{n} \,. \] Since \[\begin{align*} \left|\frac{-5-5 i}{3+3 i}\right| & = \frac{|-5-5 i|}{|3+3 i|} \\ & =\frac{\sqrt{(-5)^{2}+(-5)^{2}}}{\sqrt{3^{2}+3^{2}}} \\ & = \sqrt{\frac{50}{18}} \\ & =\frac{5}{3} > 1 \,, \end{align*}\] the above series does not converge.
Consider the series \[ \sum_{n=0}^{\infty}\left(\frac{2+i}{3-2 i}\right)^{n} \,. \] We have \[\begin{align*} \left|\frac{2+i}{3-2 i}\right| & = \frac{|2+i|}{|3-2 i|} \\ & = \frac{\sqrt{2^{2}+1^{2}}}{\sqrt{3^{2}+(-2)^{2}}} \\ & = \sqrt{\frac{5}{13}} < 1 \,. \end{align*}\] Therefore the series converges by the Geometric Series Test, and \[\begin{align*} \sum_{n=0}^{\infty}\left(\frac{2+i}{3-2 i}\right)^{n} & = \frac{1}{1-\dfrac{2+i}{3-2 i}} \\ & = \frac{1}{\dfrac{3-2 i-(2+i)}{3-2 i}} \\ & = \frac{3-2 i}{1-3 i} \\ & = \frac{3-2 i}{1-3 i} \, \frac{1+3 i}{1+3 i} \\ & = \frac{3-2 i+9 i-6 i^{2}}{1-9 i^{2}} \\ & = \frac{9+7 i}{10} \\ & = \frac{9}{10} + \frac{7}{10} i \end{align*}\]

8.3 Algebra of Limits for Series

We have proven the Algebra of Limit Theorem for sequences in \(\mathbb{C}\). A similar results holds for series as well.

Theorem 18: Algebra of Limits for Series

Let \(\left(a_{n}\right)_{n \in \mathbb{N}}\) and \(\left(b_{n}\right)_{n \in \mathbb{N}}\) be sequences in \(\mathbb{C}\) and let \(c \in \mathbb{C}\). Suppose that \[ \sum_{n=1}^{\infty} a_{n}=a \,, \qquad \sum_{n=1}^{\infty} b_{n}=b \,. \] Then:

The sum of series is the series of the sums: \[ \sum_{n=1}^{\infty}\left(a_{n} \pm b_{n}\right) = a \pm b \,. \]
The product of a series with a number obeys \[ \sum_{n=1}^{\infty} c \cdot a_{n}= c \cdot a \,. \]

Proof

Part 1. We prove the formula with the \(+\) sign, since in the other case the proof is the same. To this end, define the partial sums \[ s_{k} := \sum_{n=1}^{k} a_{n} \,, \quad t_{k} := \sum_{n=1}^{k} b_{n} \,, \quad v_{k} := \sum_{n=1}^{k}\left(a_{n}+b_{n}\right) \,. \] We can write \[\begin{align*} v_k & = \sum_{n=1}^{k} \left(a_{n}+b_{n}\right) \\ & = \left(a_{1}+b_{1}\right)+\ldots+\left(a_{k}+b_{k}\right) \\ & = \left(a_{1}+\ldots+a_{k}\right)+\left(b_{1}+\ldots+b_{k}\right) \\ & = s_{k} + t_{k} \,. \end{align*}\] By assumption \(s_{k} \to a\) and \(t_{k} \to b\). Hence, by the Algebra of Limits in \(\mathbb{C}\), we infer \[\begin{align*} \sum_{n=1}^{\infty}\left(a_{n}+b_{n}\right) & = \lim_{k \rightarrow \infty} v_{k} \\ & = \lim_{k \rightarrow \infty} \left(s_{k}+t_{k}\right) \\ & = \lim_{k \rightarrow \infty} \, s_{k} + \lim_{k \to \infty} \, t_{k} \\ & = a + b \,. \end{align*}\]

Part 2. Denote the partial sums by \[ s_{k} := \sum_{n=1}^{k} a_{n} \, , \quad t_{k} := \sum_{n=1}^{k} c \cdot a_{n} \,. \] We can write \[\begin{align*} t_k & = \sum_{n=1}^{k} c \cdot a_{k} \\ & = c \cdot a_{1} +c \cdot a_{2}+\ldots+c \cdot a_{k} \\ & = c \cdot \left(a_{1}+a_{2}+\ldots+a_{k}\right) = \\ & = c \cdot s_{k} \,. \end{align*}\] By assumption \(s_{k} \to a\), so that the Algebra of Limits in \(\mathbb{C}\) allows to conclude \[\begin{align*} \sum_{n=1}^{\infty} c \cdot a_{n} & = \lim_{k \rightarrow \infty} t_{k} \\ & = \lim_{k \rightarrow \infty} c \cdot s_{k} \\ & = c \cdot a \,. \end{align*}\]

Let us apply the Algebra of Limits for Series to a concrete example.

Example 19

Consider the following series

\[ \sum_{n=0}^{\infty}\left(2\left(\frac{1}{3}\right)^{n}+\left(\frac{2}{3}\right)^{n}\right) \,. \] Note that \[ \sum_{n=0}^{\infty}\left(\frac{1}{3}\right)^{n}=\frac{1}{1-\dfrac{1}{3}}=\frac{3}{2} \,, \quad \sum_{n=0}^{\infty}\left(\frac{2}{3}\right)^{n}=\frac{1}{1-\dfrac{2}{3}}=3 \,, \] by the Geometric Series Test of Theorem 13. Therefore we can apply the Algebra of Limit for Series to conclude \[\begin{align*} \sum_{n=0}^{\infty}\left(2\left(\frac{1}{3}\right)^{n}+\left(\frac{2}{3}\right)^{n}\right) & = 2 \cdot \sum_{n=0}^{\infty} \left(\frac{1}{3}\right)^{k} + \sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^{k} \\ & = 2 \cdot \frac{3}{2} + 3 = 6 \end{align*}\]

Important

The Algebra of Limits Theorem 18 does not discuss product and quotient of series. The situation becomes more complicated in this case: Indeed, we have \[ (a_1 + a_2) \cdot (a_2 + b_2) = a_1b_1 + a_2 b_2 + a_1 b_2 + a_2 b_1 \,. \] Therefore, in general, we can expect \[ \left( \sum_{n=0}^{\infty} a_n \right) \cdot \left( \sum_{n=0}^{\infty} b_n \right) \neq \sum_{n=0}^{\infty} a_n \cdot b_n \,. \tag{8.6}\] A way to compute \[ \left( \sum_{n=0}^{\infty} a_n \right) \cdot \left( \sum_{n=0}^{\infty} b_n \right) \] is through the so-called Cauchy Product of two series. We do not cover the latter, and the interested reader can refer to Page \(82\) in (Abbott 2015).

Similarly, we expect that \[ \dfrac{ \displaystyle\sum_{n=0}^{\infty} a_n }{ \displaystyle\sum_{n=0}^{\infty} b_n } \neq \sum_{n=0}^{\infty} \frac{a_n}{b_n} \,. \tag{8.7}\]

Let us give two examples to show that formulas (8.6) and (8.7) hold.

Example 20

We know that \[ \sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{n}=\frac{1}{1-\dfrac{1}{2}}=2 \,. \tag{8.8}\] Therefore \[ \left(\sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{n} \right) \cdot \left(\sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{n} \right) = 2 \cdot 2 = 4 \,. \] However \[ \sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{n} \cdot\left(\frac{1}{2}\right)^{n} = \sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n} = \frac{1}{1-\dfrac{1}{4}} = \frac{4}{3} \,. \] Hence \[ \left(\sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{n} \right) \cdot \left(\sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{n} \right) = 4 \neq \frac{4}{3} = \sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{n} \cdot\left(\frac{1}{2}\right)^{n} \,. \]
Using (8.8) we have \[ \frac{\displaystyle\sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{n}}{\displaystyle\sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{n}} = \frac{2}{2} = 1 \,. \] On the other hand \[ \sum_{n=0}^{\infty} \dfrac{\left(\dfrac{1}{2}\right)^{n}}{\left(\dfrac{1}{2}\right)^{n}} =\sum_{n=0}^{\infty} 1 \,, \] which does not converge by the Necessary Condition, since \(1\) does not converge to \(0\). Therefore \[ \frac{\displaystyle\sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{n}}{\displaystyle\sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{n}} \neq \sum_{n=0}^{\infty} \dfrac{\left(\dfrac{1}{2}\right)^{n}}{\left(\dfrac{1}{2}\right)^{n}} \,. \]

8.4 Non-negative series

We now investigate series of which all terms are non-negative. To be precise:

Definition 21: Non-negative series

Let \((a_{n})\) be a sequence in \(\mathbb{R}\). We call the series \[ \sum_{n=1}^{\infty} a_{n} \] a non-negative series if \[ a_n \geq 0 \,, \quad \forall \, n \in \mathbb{N}\,. \]

The key remark for non-negative series is that the partial sums are increasing.

Lemma 22

Let \((a_{n})\) be a sequence in \(\mathbb{R}\) with \[ a_n \geq 0 \,, \quad \forall \, n \in \mathbb{N}\,. \] Define the partial sums as \[ s_k := \sum_{n=1}^k a_n \,. \] The sequence \((s_k)\) is increasing.

Proof

For all \(k \in \mathbb{N}\) we have \[ s_{k+1} = \sum_{n=1}^{k+1} a_n = s_k + a_{k+1} \geq s_k \,, \] where we used that \(a_{k+1} \geq 0\). Therefore \((s_k)\) is increasing.

Therefore, if we have a series with non-negative terms \[ \sum_{n=1}^\infty a_n \] there are only \(2\) options:

\(\sum_{n=1}^\infty a_n\) converges,
\(\sum_{n=1}^\infty a_n\) diverges to \(+\infty\).

This is because the partial sums \((s_k)\) are increasing. Therefore we have either:

\((s_k)\) is bounded above: Then \(s_k\) converges by the Monotone Convergence Theorem
\((s_k)\) is not bounded above: Therefore \(s_k\) diverges to \(+\infty\).

We present several tests for the convergence of non-negative series.

8.4.1 Cauchy Condensation Test

Let us start with the study of the two non-negative series: \[ \sum_{n=1}^\infty \frac{1}{n^2} \,, \quad \sum_{n=1}^\infty \frac{1}{n} \,. \]

Question 23

Do the above series converge or diverge?

The answer is that the first series converges, while the second diverges. We prove it in the next two theorems.

Theorem 24

The following series converges \[ \sum_{n=1}^\infty \frac{1}{n^2} \,. \]

Proof

For \(k \in \mathbb{N}\) define the sequence of partial sums \[ s_{k} := \sum_{n=1}^k \frac{1}{n^2} \,. \] Note that \[\begin{align*} s_k & = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \ldots + \frac{1}{k^2} \\ & = 1 + \frac{1}{2 \cdot 2} + \frac{1}{3 \cdot 3} + \frac{1}{4 \cdot 4} + \ldots + \frac{1}{k \cdot k} \\ & < 1 + \frac{1}{2 \cdot 1} + \frac{1}{3 \cdot 2} + \frac{1}{4 \cdot 3} + \ldots + \frac{1}{k \cdot (k-1)} \\ & = 1 + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \ldots + \left( \frac{1}{k-1} - \frac{1}{k} \right) \\ & = 1 + 1 - \frac{1}{k} \\ & = 2 - \frac{1}{k} \\ & < 2 \,, \end{align*}\] showing that \(s_k\) is bounded above. Recall that \(s_k\) is increasing, by Lemma 22. Therefore, by the Monotone Convergence Theorem, we conclude that \(s_k\) converges. Hence the series \[ \sum_{n=1}^\infty \frac{1}{n^2} \] is convergent.

Theorem 25: Harmonic series

The harmonic series \[ \sum_{n=1}^\infty \frac{1}{n} \] is divergent.

Proof

For \(k \in \mathbb{N}\) define the sequence of partial sums \[ s_{k} := \sum_{n=1}^k \frac{1}{n} \,. \] Note that \[ s_2 = 1 + \frac{1}{2} \] while \[\begin{align*} s_4 & = 1 + \frac{1}{2} + \left( \frac{1}{3} + \frac{1}{4} \right) \\ & > 1 + \frac{1}{2} + \left( \frac{1}{4} + \frac{1}{4} \right) \\ & = 1 + \frac{1}{2} + 2 \left( \frac{1}{4} \right) \\ & = 1 + \frac{1}{2} + \frac{1}{2} \\ & = 1 + 2 \left( \frac{1}{2} \right) \\ \end{align*}\] Similarly \[\begin{align*} s_4 & = 1 + \frac{1}{2} + \left( \frac{1}{3} + \frac{1}{4} \right) + \left( \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right) \\ & > 1 + \frac{1}{2} + \left( \frac{1}{4} + \frac{1}{4} \right) + \left( \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} \right) \\ & = 1 + \frac{1}{2} + 2 \left( \frac{1}{4} \right) + 4 \left( \frac{1}{8} \right) \\ & = 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} \\ & = 1 + 3 \left( \frac{1}{2} \right) \\ \end{align*}\] Proceeding in a similar way, for \(k \in \mathbb{N}\) we have \[\begin{align*} s_{2^k} & = 1 + \frac{1}{2} + \left( \frac{1}{3} + \frac{1}{4} \right) + \left( \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right) \\ & \qquad + \ldots + \left( \frac{1}{2^{k-1} + 1} + \ldots + \frac{1}{2^k} \right) \\ & > 1 + \frac{1}{2} + \left( \frac{1}{4} + \frac{1}{4} \right) + \left( \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} \right) \\ & \qquad + \ldots + \left( \frac{1}{2^k} + \ldots + \frac{1}{2^k} \right) \\ & = 1 + \frac{1}{2} + 2 \left( \frac{1}{4} \right) + 4 \left( \frac{1}{8} \right) + \ldots + 2^{k-1} \left( \frac{1}{2^k} \right)\\ & = 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \ldots + \frac{1}{2} \\ & = 1 + k \left( \frac{1}{2} \right) \\ \end{align*}\] Hence \[ s_{2^k} > 1 + k \left( \frac{1}{2} \right) \,, \quad \forall \, k \in \mathbb{N}\,, \] showing that \(s_{2^k}\) is unbounded. Therefore \(s_k\) is unbounded, and \(s_k\) does not converge. Therefore the series \[ \sum_{n=1}^\infty \frac1n \] is divergent.

The proofs of the above theorems inspire the Cauchy Condensation Test.

Theorem 26: Cauchy Condensation Test

Let \((a_n)\) be a sequence in \(\mathbb{R}\). Suppose that \((a_n)\) is non-negative and decreasing, that is, \[ a_{n} \geq a_{n+1} \,, \quad \forall \, n \in \mathbb{N}\,. \] They are equivalent:

The series \[ \sum_{n=1}^\infty a_n \] converges.
The series \[ \sum_{n=0}^\infty 2^n a_{2^n} = a_1 + 2a_2 + 8a_8 + 16 a_{16} + \ldots \] converges.

Proof

For \(k \in \mathbb{N}\) denote the partial sums by \[ s_k := \sum_{n=1}^k a_n \,, \quad t_k := \sum_{n=0}^k 2^n a_{2^n} \,. \] Since \(a_n \geq 0\), it is immediate to check that the sequences \((s_k)\) and \((t_k)\) are increasing.

Part 1. Assume that the series \[ \sum_{n=0}^\infty 2^n a_{2^n} \] diverges. Hence the sequence \((t_k)\) diverges and therefore \((t_k)\) is not bounded above.

Indeed, suppose \((t_k)\) was bounded above. Since \((t_k)\) is increasing, we would conclude that \((t_k)\) is convergent by the Monotone Convergence Theorem. Contradiction.

We want to estimate \(s_k\) from below. To this end, we notice that \[\begin{align*} s_2 & = a_1 + a_2 \\ & \geq \frac{1}{2} a_1 + a_2 \\ & = \frac{1}{2} \left( a_1 + 2a_2 \right) \\ & = \frac{1}{2} \, t_1 \,, \end{align*}\] where we used that \(a_n \geq 0\), and so \(a_1 > a_1/2\). Moreover \[\begin{align*} s_4 & = a_1 + a_2 + (a_3 + a_4) \\ & \geq a_1 + a_2 + (a_4 + a_4) \\ & = a_1 + a_2 + 2a_4 \\ & \geq \frac{1}{2} \, a_1 + a_2 + 2 a_4 \\ & = \frac{1}{2} \left( a_1 + 2a_2 + 4a_4 \right) \\ & = \frac{1}{2} \, t_2 \,, \end{align*}\] where we used also that \((a_n)\) is decreasing. Similar reasoning yields \[\begin{align*} s_8 & = a_1 + a_2 + (a_3 + a_4) + (a_5 + a_6 + a_7 + a_8) \\ & \geq a_1 + a_2 + (a_4 + a_4) + (a_8 + a_8 + a_8 + a_8) \\ & = a_1 + a_2 + 2 a_4 + 4 a_8 \\ & \geq \frac{1}{2} \, a_1 + a_2 + 2 a_4 + 4 a_8 \\ & = \frac{1}{2} \left( a_1 + 2a_2 + 4a_4 + 8 a_8 \right) \\ & = \frac{1}{2} \, t_3 \,. \end{align*}\] where again we used that \((a_n)\) is decreasing, and \(a_n \geq 0\). Iterating, we obtain that for all \(k \in \mathbb{N}\) it holds: \[\begin{align*} s_{2^k} & = a_1 + a_2 + (a_3 + a_4) + (a_5 + a_6 + a_7 + a_8) \\ & \qquad + \ldots + (a_{2^{k-1}} + \ldots + a_{2^k}) \\ & \geq a_1 + a_2 + (a_4 + a_4) + (a_8 + a_8 + a_8 + a_8) \\ & \qquad + \ldots + (a_{2^k} + \ldots + a_{2^k}) \\ & = a_1 + a_2 + 2 a_4 + 4 a_8 + \ldots + 2^{k-1} a_{2^k}\\ & \geq \frac{1}{2} \, a_1 + a_2 + 2 a_4 + 4 a_8 + \ldots + 2^{k-1} a_{2^k} \\ & = \frac{1}{2} \left( a_1 + 2a_2 + 4a_4 + 8 a_8 + \ldots + 2^k a_{2^k} \right) \\ & = \frac{1}{2} \, t_k \,. \end{align*}\] We have shown that \[ s_{2^k} \geq \frac{1}{2} \, t_k \,, \quad \forall \, k \in \mathbb{N}\,. \] Since \((t_k)\) is not bounded above, we infer that \((s_{2^k})\) is not bounded above. In particular \((s_k)\) is not bounded, and hence divergent. Thus the series \[ \sum_{n=1}^\infty a_n \] diverges.

Part2. Suppose that the series \[ \sum_{n=0}^\infty 2^n a_{2^n} \] converges. Hence the sequence \((t_k)\) converges, and therefore \((t_k)\) is bounded. This means that there exists \(M > 0\) such that \[ |t_k| \leq M \,, \quad \forall \, n \in \mathbb{N}\,. \] Since \(t_k \geq 0\), the above reads \[ t_k \leq M \,, \quad \forall \, n \in \mathbb{N}\,. \] Fix \(k \in \mathbb{N}\) and let \(m \in \mathbb{N}\) be such that \[ k \leq 2^{m+1} - 1 \,. \] In this way \[ s_k \leq s_{2^{m+1}-1} \,. \] We have \[\begin{align*} s_{2^{m+1}-1} & = a_1 + (a_2 + a_3) + (a_4 + a_5 + a_6 + a_7) \\ & \qquad + \ldots + (a_{2^m} + \ldots + a_{2^{m+1}-1}) \\ & \leq a_1 + (a_2 + a_2) + (a_4 + a_4 + a_4 + a_4) \\ & \qquad + \ldots + (a_{2^m} + \ldots + a_{2^m} ) \\ & = a_1 + 2a_2 + 4a_4 + \ldots + 2^m a_{2^m} \\ & = t_m \,, \end{align*}\] where we used that \((a_n)\) is decreasing. We have then shown \[ s_k \leq s_{2^{m+1}-1} \leq t_m \leq M \,. \] Since \(M\) does not depend on \(k\), we conclude that \[ s_k \leq M \,, \quad \forall \, k \in \mathbb{N}\,. \] As \(s_k \geq 0\), we conclude that \(s_k\) is bounded. Recalling that \((s_k)\) is increasing, by the Monotone Convergence Theorem we infer that \((s_k)\) converges. This proves that the series \[ \sum_{n=1}^\infty a_n \] is convergent, ending the proof.

Thanks to the Cauchy Condensation Test of Theorem 26 we can prove the following result.

Theorem 27: Convergence of \(p\)-series

Let \(p \in \mathbb{R}\). Consider the \(p\)-series \[ \sum_{n=1}^{\infty} \frac{1}{n^{p}} \,. \] We have:

If \(p>1\) the \(p\)-series converges.
If \(p \leq 1\) the \(p\)-series diverges.

Proof

The series in question is \[ \sum_{n=1}^\infty a_n \,, \quad a_n := \frac{1}{n^p} \,. \] Note that \((a_n)\) is decreasing and non-negative. Hence, by the Cauchy Condensation Test of Theorem 26, the \(p\)-series converges if and only if \[ \sum_{n=0}^\infty 2^n a_{2^n} \] converges. We have \[ \sum_{n=0}^\infty 2^n a_{2^n} = \sum_{n=0}^\infty 2^{n-np} = \sum_{n=0}^\infty (2^{1-p})^{n} \,, \] and the latter is a Geometric Series of ratio \[ x := 2^{1-p} \,. \] By the Geometric Series Test, we have convergence of \[ \sum_{n=0}^\infty (2^{1-p})^{n} \] if and only if \[ |x| < 1 \,. \] The above is equivalent to \[ 2^{1-p} < 1 = 2^0 \quad \iff \quad 1 - p < 0 \quad \iff \quad p > 1 \,. \] Therefore \[ \sum_{n=1}^\infty \frac{1}{n^p} \] converges if and only if \(p>1\), ending the proof.

The following is another notable application of the Cauchy Condensation Test.

Theorem 28

Let \(p \in \mathbb{R}\). Consider the series \[ \sum_{n=2}^{\infty} \frac{1}{n \left( \log n \right)^p} \,. \] We have:

If \(p>1\) the series converges.
If \(p \leq 1\) the series diverges.

Proof

The series in question is \[ \sum_{n=2}^{\infty} a_n \,, \quad a_n := \frac{1}{n \left( \log n \right)^p} \,. \] Note that \((a_n)\) is non-negative and decreasing. Therefore we can apply the Cauchy Condensation Test to conclude that the above series is convergent if and only if the series \[ \sum_{n=1}^\infty 2^n a_{2^n} \] is convergent. We have \[ 2^n a_{2^n} = 2^n \, \frac{1}{2^n \left( \log 2^n \right)^p } = \frac{1}{n^p \, \log 2} \] so that \[ \sum_{n=1}^\infty 2^n a_{2^n} = \frac{1}{\log 2 } \, \sum_{n=1}^\infty \frac{1}{n^p} \,. \] The latter is a \(p\)-series, which by Theorem 27 converges if and only if \(p > 1\). Hence \[ \sum_{n=2}^{\infty} \frac{1}{n \left( \log n \right)^p} \] converges if and only if \(p > 1\), and the proof is concluded.

8.4.2 Comparison Test

Another really useful result to study the convergence of non-negative series is the Comparison Test.

Theorem 29: Comparison test

Let \(\left(a_{n}\right)_{n \in \mathbb{N}}\) and \(\left(b_{n}\right)_{n \in \mathbb{N}}\) be non-negative sequences. Suppose that there exists \(N \in \mathbb{N}\) such that \[ a_{n} \leq b_{n} \,, \quad \, \forall \, n \geq N \,. \] They hold:

If \(\sum_{n=1}^{\infty} b_{n}\) converges, then also \(\sum_{n=1}^{\infty} a_{n}\) converges.
If \(\sum_{n=1}^{\infty} a_{n}\) diverges, then also \(\sum_{n=1}^{\infty} b_{n}\) diverges.

Proof

Part 1. Define the partial sums starting at \(n=N\) \[ s_k := \sum_{n=N}^k a_k \,, \quad t_k := \sum_{n=N}^k b_k \,. \] Suppose that \[ \sum_{n=1}^{\infty} b_{n} \] converges. Hence also the series \[ \sum_{n=N}^{\infty} b_{n} \] converges. Then \((t_k)\) is a convergent sequence, which implies that \((t_k)\) is bounded, and hence bounded above. We have that \(s_k\) is bounded above: Indeed, using the assumption, we have \[\begin{align*} s_k & = a_N + a_{N+1} + \ldots + a_k \\ & \leq b_N + b_{N+1} + \ldots + b_k \\ & = t_k \,, \end{align*}\] which reads \[ s_k \leq t_k \,, \quad \forall \, k \geq N \,. \] Therefore \((s_k)\) is bounded above, being \((t_k)\) bounded above. Recall that \(s_k\) is increasing, by Lemma 22. By the Monotone Convergence Theorem we conclude that \(s_k\) is convergent, showing that the series \[ \sum_{n=N}^\infty a_n \] converges. Hence also the series \[ \sum_{n=1}^\infty a_n \] converges, concluding the proof of Point 1.

Part 2. Note that Point 2 is the contrapositive of Point 1, and hence it holds.

Let us give two applications of the Comparison Test.

Example 30

Consider the series \[ \sum_{n=1}^{\infty} \frac{1}{n^{2}+3 n-1} \,. \tag{8.9}\] Since \(3 n-1 \geq 0\) for all \(n \in \mathbb{N}\), we get \[ \frac{1}{n^{2}+3 n-1} \leq \frac{1}{n^{2}} \,, \quad \forall \, n \in \mathbb{N}\,. \] By Theorem 27 the \(p\)-series \[ \sum_{n=1}^{\infty} \frac{1}{n^{2}} \] converges. Therefore also the series at (8.9) converges by the Comparison Test in Theorem 29.
Consider the series \[ \sum_{n=0}^{\infty} \frac{3^{n}+6 n+\dfrac{1}{n+1}}{2^{n}} \,. \tag{8.10}\] Note that \[ \frac{3^{n}+6 n+\dfrac{1}{n+1}}{2^{n}} \geq \frac{3^{n}}{2^{n}}=\left(\frac{3}{2}\right)^{n} \,, \quad \forall \, n \in \mathbb{N}\,. \] Since \(\left|\frac{3}{2}\right|=\frac{3}{2}>1\), the series \[ \sum_{n=0}^{\infty}\left(\frac{3}{2}\right)^{n} \] diverges by the Geometric Series Test in Theorem 13. Therefore, by the Comparison Test, also the series at (8.10) diverges.

8.4.3 Limit Comparison Test

To apply the Comparison Test to the series \[ \sum_{n=1}^\infty a_n \,, \] one needs to find another sequence \((b_n)\) such that \[ a_n \leq b_n \,,\quad \forall \, n \geq N \,, \] or \[ b_n \leq a_n \,, \quad \forall \, n \geq N \,. \] This is not always possible. However, one might be able to show that \[ L = \lim_{n\to \infty} \frac{a_n}{b_n} \] for some \(L \in \mathbb{R}\). In this case, the series of \((a_n)\) and \((b_n)\) can still be compared, in the sense specified in the below theorem.

Theorem 31: Limit Comparison Test

Let \((a_n)\) and \((b_n)\) be sequences such that \[ a_n \geq 0 \,, \quad b_n > 0 \,, \quad \forall \, n \in \mathbb{N}\,. \] Suppose there exists \(L \in \mathbb{R}\) such that \[ L = \lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}} \,. \] They hold:

If \(0<L<\infty\), then \[ \sum_{n=1}^{\infty} a_{n} \,\, \text { converges } \Longleftrightarrow \quad \sum_{n=1}^{\infty} b_{n} \,\, \text { converges. } \]
If \(L=0\), then
- If \(\sum_{n=1}^{\infty} b_{n}\) converges also \(\sum_{n=1}^{\infty} a_{n}\) converges,
- If \(\sum_{n=1}^{\infty} a_{n}\) diverges also \(\sum_{n=1}^{\infty} b_{n}\) diverges.

Proof

Part 1. Suppose that \(0<L<1\). Set \[ \varepsilon:= \frac{L}{2} \,. \] Since \(\varepsilon>0\) and \(a_n/b_n \to L\), there exists \(N \in \mathbb{N}\) such that \[ \left|\frac{a_{n}}{b_{n}} - L \right| < \varepsilon\,, \quad \forall \, n \geq N \,. \] The above is equivalent to \[ L - \varepsilon< \frac{a_{n}}{b_{n}} < \varepsilon+ L \,, \quad \forall \, n \geq N \,. \] Since \(\varepsilon= L/2\), we get \[ \frac{L}{2} < \frac{a_{n}}{b_{n}} < \frac{3L}{2} \,, \quad \forall \, n \geq N \,, \] or equivalently, \[ \frac{L}{2} \, b_n < a_n < \frac{3L}{2} \, b_n \,, \quad \forall \, n \geq N \,. \] We are now ready to prove the main claim:

Suppose that \[ \sum_{n=1}^{\infty} a_{n} \] converges. Then also \[ \sum_{n=N}^{\infty} a_{n} \,, \] since we are only discarding a finite number of terms. As \[ \frac{L}{2} b_{n} \leq a_{n} \,, \quad \forall \, n \geq N \,, \] it follows from the Comparison Test in Theorem 29 that the series \[ \sum_{n=N}^{\infty} \frac{L}{2} b_{n} \,. \] converges. Since \(L/2\) is a constant, we also conclude that \[ \sum_{n=1}^{\infty} b_{n} \] converges.
Suppose that \[ \sum_{n=1}^{\infty} b_{n} \] converges. Then also \[ \sum_{n=N}^{\infty} \frac{3 L}{2} b_{n} \] converges. Since \[ a_{n} < \frac{3 L}{2} b_{n} \,, \quad \forall \, n \geq N \,, \] by the Comparison Test we infer that \[ \sum_{n=N}^{\infty} a_{n} \] converges. Therefore, also \[ \sum_{n=1}^{\infty} a_{n} \] converges.

Part 2. Suppose that \(L = 0\). Note that the second bullet point is the contrapositive of the first. Hence we only need to show the first bullet point. Let \(\varepsilon= 1\). Since \(a_n / b_n \to 0\), there exists \(N \in \mathbb{N}\) such that \[ \left|\frac{a_{n}}{b_{n}}-0\right| < \varepsilon= 1 \,, \quad \forall \, n \geq N \,. \] Therefore \[ a_{n} < b_{n} \,, \quad \forall \, n \geq N \,. \] The thesis follows immediately by the Comparison Test in Theorem 29.

Important

It might happen that \[ \lim_{n \to \infty } \, \frac{a_{n}}{b_{n}} = \infty \,. \] In this case it can be helpful to interchange the roles of \(a_{n}\) and \(b_{n}\), since then \[ \lim_{n \to \infty } \, \frac{b_{n}}{a_{n}} = 0 \,. \]

Let us give a few applications of the Limit Comparison Tets.

Example 32

The following series converges \[ \sum_{n=1}^{\infty} \frac{2 n^{3}+5 n+1}{7 n^{6}+2 n+5} \,. \]

Proof. Set \[ a_n := \frac{2 n^{3}+5 n+1}{7 n^{6}+2 n+5}\,, \quad b_n := \frac{1}{n^3} \,. \] We have \[\begin{align*} L & :=\lim_{n \rightarrow \infty} \frac{a_n}{b_n} \\ & = \lim_{n \to \infty} \frac{2 n^{3}+5 n+1}{7 n^{6}+2 n+5} \bigg/ \frac{1}{n^{3}} \\ & = \lim_{n \rightarrow \infty} \frac{2 n^{6}+5 n^{4}+n^{3}}{7 n^{6}+2 n+5} \\ & = \lim_{n \rightarrow \infty} \frac{2+ \dfrac{5}{n^{2}} + \dfrac{1}{n^{3}} }{7 + \dfrac{2}{n^{5}}+ \dfrac{5}{n^{6}} } = \frac{2}{7} \,. \end{align*}\] The series \[ \sum_{n=1}^{\infty} \frac{1}{n^{3}} \] converges, being a \(p\)-series with \(p =3 > 1\). Since \(L = \frac{2}{7}>0\), also the series of interest converges, by the Limit Comparison Test.

Example 33

Prove that the following series diverges \[ \sum_{n=1}^{\infty} \frac{n+\cos (n)}{n^{2}} \,. \]

Proof. We expect the terms in the series to behave like \(1/n\) for large \(n\). Hence we set \[ a_n := \frac{n+\cos (n)}{n^{2}} \,, \quad b_n = \frac{1}{n} \,. \] We compute \[\begin{align*} L & := \frac{a_n}{b_n} \\ & = \lim_{n \rightarrow \infty} \frac{n+\cos (n)}{n^{2}} \bigg/ \frac{1}{n} \\ & = \lim_{n \rightarrow \infty} \frac{n^{2}+n \cos (n)}{n^{2}} \\ & = \lim_{n \rightarrow \infty} \left(1+\frac{\cos (n)}{n}\right) \end{align*}\] Since \[ -1 \leq \cos (n) \leq 1 \,, \] we obtain \[ -\frac{1}{n} \leq \frac{\cos (n)}{n} \leq \frac{1}{n} \,. \] As both \(-\frac{1}{n} \rightarrow 0\) and \(\frac{1}{n} \rightarrow 0\), by the Squeeze Theorem \[ \frac{\cos (n)}{n} \longrightarrow 0 \,. \] Hence \[ L = \lim_{n \rightarrow \infty} \left(1+\frac{\cos (n)}{n}\right) = 1 \,. \] The harmonic series \[ \sum_{n=1}^{\infty} \frac{1}{n} \] does not converge. Since \(L = 1 > 0\), the series of interest diverges by the Limit Comparison Test.

Example 34

Prove that the series \[ \sum_{n=1}^\infty \left( 1 - \cos\left( \frac{1}{n} \right) \right) \] converges.

Solution. Since \[ \cos \left( \frac{1}{n} \right) \leq 1 \,, \] the above is a non-negative series. Recall the limit \[ \lim_{n \to \infty} \frac{1 - \cos(a_n)}{(a_n)^2} = \frac{1}{2} \,, \] where \((a_n)\) is a sequence in \(\mathbb{R}\) such that \(a_n \to 0\) and \[ a_n \neq 0 \quad \forall \, n \in \mathbb{N}\,. \] In particular, for \(a_n = 1/n\), we obtain \[ \lim_{n \to \infty} \, n^2 \, \left(1 - \cos \left( \frac{1}{n} \right) \right) = \frac{1}{2} \,. \] Set \[ b_n : = 1 - \cos\left( \frac{1}{n} \right) \,, \quad c_n := \frac{1}{n^2} \,. \] We have \[\begin{align*} L & := \lim_{n \to \infty} \frac{b_n}{c_n} \\ & = \lim_{n \to \infty} n^2 \, \left(1 - \cos \left( \frac{1}{n} \right) \right) \\ & = \frac{1}{2} \,. \end{align*}\] Note that the series \[ \sum_{n=1}^\infty \frac{1}{n^2} \] converges, being a \(p\)-series with \(p>2\). Therefore, since \(L = 1/2 >0\), also the series \[ \sum_{n=1}^\infty \left( 1 - \cos\left( \frac{1}{n} \right) \right) \] converges, by the Limit Comparison Test.

Sometimes the Limit Comparison Test fails, but the Comparison Test works.

Example 35

Consider the series \[ \sum_{n=1}^{\infty} \frac{1+\sin (n)}{n^{2}} \,. \] Since \[ \sin (n) \geq-1 \,, \] the above is a non-negative series. We expect this series to behave similar to \[ \sum_{n=1}^{\infty} \frac{1}{n^{2}} \,. \] However \[ \frac{1+\sin (n)}{n^{2}} \bigg/ \frac{1}{n^{2}}=1+\sin (n) \] does not converge. Hence, we cannot use the Limit Comparison Test. In alternative, we note that \[ \frac{1+\sin (n)}{n^{2}} \leq \frac{2}{n^{2}} \,, \quad \forall \, n \in \mathbb{N}\,. \] The series \[ \sum_{n=1}^{\infty} \frac{2}{n^{2}} \] converges, being a \(p\)-series with \(p=2>1\). Therefore also \[ \sum_{n=1}^{\infty} \frac{1+\sin (n)}{n^{2}} \] converges, by the Comparison Test of Theorem 29.

8.4.4 Ratio Test for positive series

The Ratio Test can be generalized to series. Notice that in this case the terms of the series need to be positive.

Theorem 36: Ratio Test for positive series

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

Suppose that the following limit exists: \[ L:=\lim_{n \to \infty} \frac{a_{n+1}}{a_{n}} \,. \] They hold:
- If \(L<1\) then \(\sum_{n=1}^\infty a_n\) converges.
- If \(L>1\) then \(\sum_{n=1}^\infty a_n\) diverges.
Suppose that there exists \(N \in \mathbb{N}\) and \(L>1\) such that \[ \frac{a_{n+1}}{a_{n}} \geq L \,, \quad \forall \, n \geq N \,. \] Then the series \(\sum_{n=1}^\infty a_n\) diverges.

Proof

Part 1. Let \[ L:=\lim_{n \to \infty} \frac{a_{n+1}}{a_{n}} \,. \]

Suppose that \(L<1\). Therefore there exists \(r\) such that \[ L < r < 1 \,. \] Define \[ \varepsilon:= r - L \,, \] so that \(\varepsilon>0\). By the convergence \(a_{n+1}/a_n \to L\) there exists \(N \in \mathbb{N}\) such that \[ \left|\frac{a_{n+1}}{a_{n}}-L\right| < \varepsilon= r - L \,, \quad \forall \, n \geq N \,. \] In particular \[ \frac{a_{n+1}}{a_{n}}-L < r - L \,, \quad \forall \, n \geq N \,, \] which implies \[ a_{n+1} < r \, a_n \,, \quad \forall \, n \geq N \,. \] Applying \(n-N\) times the above estimate we get \[ 0 < a_{n} < r \, a_{n-1} < \ldots < r^{n-N} \, a_{N} \,, \quad \forall \, n \geq N \,. \tag{8.11}\] Note that the series of \(r^{n-N} \, a_{N}\) converges, since \[ \sum_{n=N}^\infty r^{n-N} \, a_{N} = a_{N} \sum_{k=0}^\infty r^k = a_N \frac{1}{1-r} \,, \] where the last equality follows because \(\sum_{k=0}^\infty r^k\) is a geometric series and \(0<r<1\). Since (8.11) holds, by the Comparison Test in Theorem 29 we conclude that the series \[ \sum_{n=N}^\infty a_n \] converges. Therefore also the series \[ \sum_{n=1}^\infty a_n \] converges, ending the proof in the case \(L<1\).
Suppose \(L>1\): Then, by the Ratio Test for sequences, it follows that \(a_n\) diverges. Therefore \[ \lim_{n \rightarrow \infty} a_{n} \neq 0 \,, \] and the series \(\sum_{n=1}^\infty\) diverges by the Necessary Condition, see Theorem 6.

Part 2. Suppose there exists \(L>1\) and \(N \in \mathbb{N}\) such that \[ \frac{a_{n+1}}{a_n} \geq L \,, \quad \forall \, n \geq N \,. \] By the Ratio Test for sequences it follows that \(a_n\) diverges. Therefore \(\sum_{n=1}^\infty\) diverges by the Necessary Condition.

Example 37

Consider the series \[ \sum_{n=1}^{\infty} \frac{(n !)^{2}}{(2 n) !} \] Let \[ a_{n}=\frac{(n !)^{2}}{(2 n) !} \] and compute \[\begin{align*} \frac{a_{n+1}}{a_{n}} & =\frac{((n+1) !)^{2}}{(2(n+1)) !} \bigg/ \frac{(n !)^{2}}{(2 n) !} \\ & = \frac{((n+1) !)^{2}}{(n !)^{2}} \frac{(2 n) !}{(2(n+1)) !} \\ & = \left(\frac{(n+1) !}{n !}\right)^{2} \frac{(2 n) !}{(2 n+2) !} \\ & = \frac{(n+1)^{2}}{(2 n+2)(2 n+1)} \\ & = \frac{\left(1+\dfrac{1}{n}\right)^{2}}{\left(2+\dfrac{2}{n}\right)\left(2+\dfrac{1}{n}\right)} \end{align*}\] Therefore \[ L = \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{\left(1+\dfrac{1}{n}\right)^{2}}{\left(2+\dfrac{2}{n}\right)\left(2+\dfrac{1}{n}\right)} = \frac{1}{4} \,. \] Since \(L = 1/4 <1\), be the Ratio Test for Series we conclude that the series of interest converges.

Important

Like with the Ratio Test for sequences, the case \(L=1\) is not covered by Theorem 36. This is because, in this case, the series \[ \sum_{n=1}^\infty a_n \] might be convergent or divergent, as shown in the next example.

Example 38

Consider the series \[ \sum_{n=1}^\infty \frac{1}{n} \,. \] Setting \(a_n = 1/n\), we have \[\begin{align*} L & = \lim_{n\to \infty} \frac{a_{n+1}}{a_n} \\ & = \lim_{n \to \infty} \frac{1}{n+1} \bigg/ \frac{1}{n} \\ & = \lim_{n \to \infty} \frac{n}{n+1} = 1 \,. \end{align*}\] Therefore \(L = 1\) and we cannot apply the Ratio Test. However the series in question diverges, being the harmonic series.
Consider the series \[ \sum_{n=1}^\infty \frac{1}{n^2} \,. \] Setting \(a_n = 1/n^2\), we have \[\begin{align*} L & = \lim_{n\to \infty} \frac{a_{n+1}}{a_n} \\ & = \lim_{n \to \infty} \frac{1}{(n+1)^2} \bigg/ \frac{1}{n^2} \\ & = \lim_{n \to \infty} \frac{n^2}{n^2 + 2n +1} = 1 \,. \end{align*}\] Therefore \(L=1\) and we cannot apply the Ratio Test. However the series in question diverges, being a \(p\)-series with \(p = 2 >1\).

The Ratio Test can often be combined with other convergence tests, as seen in the following example.

Example 39

Using the Cauchy Condensation Test and the Ratio Test, prove that the series below converges \[ \sum_{n=1}^\infty \frac{\log (n)}{ n^{2} } \,. \]

Solution. With \(a_{n}=\log n / n^{2}\), we have \[ \sum_{n=0}^{\infty} 2^{n} a_{2^{n}} =\sum_{n=2}^{\infty} 2^{n} \frac{\log (2^{n})}{\left(2^{n}\right)^{2}} =\log (2) \, \sum_{n=2}^{\infty} \frac{n }{2^{n}} \] This converges according to the Ratio Test: With \(b_{n}=n / 2^{n}\), we have \[ \frac{b_{n+1}}{b_n} = \frac{n+1}{2^{n+1}} \bigg/ \frac{n}{2^n} = \frac{n+1}{2n} \longrightarrow \frac12 < 1 \,. \] Hence \(\sum_{n=1}^{\infty} \frac{\log n}{n^{2}}\) converges by the Cauchy Condensation Test.

8.5 General series

In the previous section we presented several tests for non-negative series. For non-negative series we shows that the partial sums \((s_k)\) are increasing, see Lemma 22. This makes non-negative terms series relatively easy to study.

When a series contains both positive and negative terms, the partial sums \((s_k)\) might oscillate, making the series harder to study. In this section we present some tests for general series in \(\mathbb{C}\).

8.5.1 Absolute Convergence Test

To study general series, we introduce a stronger notions of convergence, known as absolute convergence.

Definition 40: Absolute convergence

Let \(\left(a_{n}\right)\) be a sequence in \(\mathbb{C}\). The series \(\sum_{n=1}^{\infty} a_{n}\) is said to converge absolutely if the non-negative series \[ \sum_{n=1}^{\infty}\left|a_{n}\right| \] converges.

Let us show that absolute convergence implies convergence.

Theorem 41: Absolute Convergence Test

Let \(\left(a_{n}\right)\) be a sequence in \(\mathbb{C}\). If the series \(\sum_{n=1}^{\infty} a_{n}\) converge absolutely, then the series converges.

Before proceeding with the proof, we introduce some notation.

Notation 42: Positive and negative part

For a number \(x \in \mathbb{R}\), we define \[ x^{+}:= \begin{cases} x & \mbox{ if } \, x \geq 0 \\ 0 & \mbox{ if } \, x < 0 \\ \end{cases} \qquad x^{-}:= \begin{cases} -x & \mbox{ if } \, x < 0 \\ 0 & \mbox{ if } \, x \geq 0 \\ \end{cases} \] These are called the positive part and negative part of \(x\), respectively. Note that \[ x^+ \geq 0 \,, \quad x^{-} \geq 0 \,. \] Moreover they hold \[ x = x^+ - x^- \,, \quad |x| = x^+ + x^- \,. \] The above relations are easy to check, and the proof is omitted. In particular, it holds that \[ x^+ \leq |x| \,, \quad x^- \leq |x| \,. \]

Proof: Proof of Theorem 41

Part 1. Suppose first that \((a_n)\) is a sequence in \(\mathbb{R}\) such that \[ \sum_{n=1}^\infty |a_n| \] converges. Since \[ 0 \leq a_{n}^{+} \leq\left|a_{n}\right| \,, \quad \forall \, n \in \mathbb{N}\,, \] we can use the Comparison Test for non-negative series (Theorem 29) and conclude that the series \[ \sum_{n=1}^{\infty} a_{n}^{+} \,. \] converges. Similarly, we have \[ 0 \leq a_{n}^{-} \leq\left|a_{n}\right| \,, \quad \forall \, n \in \mathbb{N}\,. \] Therefore also the series \[ \sum_{n=1}^{\infty} a_{n}^{-} \] converges by the Comparison Test. Since \[ a_n = a_n^+ - a_n^- \,, \quad \forall \, n \in \mathbb{N}\,, \] we can use the Algebra of Limits for series (Theorem 18) to conclude that \[ \sum_{n=1}^{\infty} a_{n} =\sum_{n=1}^{\infty}\left(a_{n}^{+}-a_{n}^{-}\right)=\sum_{n=1}^{\infty} a_{n}^{+}-\sum_{n=1}^{\infty} a_{n}^{-} \] converges.

Part 2. Suppose now that \((a_n)\) is a sequence in \(\mathbb{C}\) such that \[ \sum_{n=1}^\infty |a_n| \] converges. Let \(x_n, y_n \in \mathbb{R}\) denote the real and imaginary part of \(a_n\). Therefore \[ |x_n| = \sqrt{x_n^2} \leq \sqrt{x_n^2 + y_n^2} = |a_n| \,, \quad \forall \, n \in \mathbb{N}\,. \] Therefore the series \[ \sum_{n=1}^\infty |x_n| \] converges by the Comparison Test for non-negative series (Theorem 29). Since \((x_n)\) is a real sequence, from Part 1 of the proof we have that the series \[ \sum_{n=1}^\infty x_n \] converges. Arguing in the same way for the imaginary part \(y_n\) we conclude that also \[ \sum_{n=1}^\infty y_n \] converges. Finally, by the Algebra of Limits in \(\mathbb{C}\), we get \[ \sum_{n=1}^\infty a_n = \sum_{n=1}^\infty \left( x_n + i y_n \right) = \sum_{n=1}^\infty x_n + i \sum_{n=1}^\infty y_n \,, \] proving that \(\sum_{n=1}^\infty a_n\) converges.

Example 43

The series \[ \sum_{n=1}^{\infty}(-1)^{n} \frac{1}{n} \] does not converge absolutely, since \[ \sum_{n=1}^{\infty}\left|(-1)^{n} \frac{1}{n}\right|=\sum_{n=1}^{\infty} \frac{1}{n} \] doesn’t converge, being the harmonic series.

Example 44

Prove that the following series converges \[ \sum_{n=1}^{\infty}(-1)^{n} \frac{n^{2}-5 n+2}{n^{4}} \,. \]

Solution. We have \[ \left|(-1)^{n} \frac{n^{2}-5 n+2}{n^{4}}\right|=\frac{\left|n^{2}-5 n+2\right|}{n^{4}} \leq \frac{n^{2}+5 n+2}{n^{4}} \] where we used the triangle inequality. Note that \[\begin{align*} \frac{n^{2}+5 n+2}{n^{4}} \bigg/ \frac{1}{n^{2}} & =\frac{n^{4}+5 n^{3}+2 n^{2}}{n^{4}} \\ & = 1+ \frac{5}{n} + \frac{2}{n^{2}} \longrightarrow 1 \end{align*}\] The series \[ \sum_{n=1}^{\infty} \frac{1}{n^{2}} \] converges, being a \(p\)-series with \(p=2\). Hence, also \[ \sum_{n=1}^{\infty} \frac{n^{2}+5 n+2}{n^{4}} \] converges by the Limit Comparison Test for non-negative series (Theorem 31). Since \[\begin{align*} 0 \leq \left|(-1)^{n} \frac{n^{2}-5 n+2}{n^{4}}\right| & = \frac{\left|n^{2}-5 n+2\right|}{n^{4}} \\ & \leq \frac{n^{2}+5 n+2}{n^{4}} \,, \end{align*}\] the series \[ \sum_{n=1}^{\infty}\left|(-1)^{n} \frac{n^{2}-5 n+2}{n^{4}}\right| \] converges by the Comparison Test for non-negative series (Theorem 29). This shows the series \[ \sum_{n=1}^{\infty}(-1)^{n} \frac{n^{2}-5 n+2}{n^{4}} \] converges by the Absolute Convergence Test.

8.5.2 Ratio Test for general series

As an application of the Absolute Convergence Test we obtain the Ratio Test for general series.

Theorem 45: Ratio Test for general series

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

Suppose that the following limit exists: \[ L:=\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_{n}}\right| \,. \] They hold:
- If \(L<1\) then \(\sum_{n=1}^\infty a_n\) converges absolutely, and hence converges.
- If \(L>1\) then \(\sum_{n=1}^\infty a_n\) diverges.
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 series \(\sum_{n=1}^\infty a_n\) diverges.

Proof

Part 1. Let \[ b_{n} := \left|a_{n}\right| \,, \] so that \[ L:= \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_{n}}\right| = \lim_{n \to \infty} \frac{b_{n+1}}{b_{n}} \,. \]

Suppose that \(L<1\). Since \((b_n)\) is a sequence with non-negative terms, we have that \[ \sum_{n=1}^{\infty} b_{n} \] converges by the Ratio Test for non-negative series, see Theorem 36. Since, by definition \[ \sum_{n=1}^{\infty} b_{n}=\sum_{n=1}^{\infty}\left|a_{n}\right| \] also the latter series converges, i.e., \(\sum_{n=1}^{\infty} a_{n}\) converges absolutely. In particular \(\sum_{n=1}^{\infty} a_{n}\) converges, by the Absolute Convergence Test in Theorem 41.
Suppose that \(L>1\). Then the sequence \((a_n)\) diverges by the Ratio Test for sequences. Hence the series \[ \sum_{n=1}^\infty a_n \] diverges by the Necessary Condition in Theorem 6.

Part 2. If 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 \((a_n)\) diverges by the Ratio Test for sequences, and we conclude as above.

Example 46

Prove that the below series converges \[ \sum_{n=1}^{\infty} \frac{(4-3 i)^{n}}{(n+1) !} \]

Solution. Set \[ a_{n} := \frac{(4-3 i)^{n}}{(n+1) !} \,. \] Then \[\begin{align*} L & := \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \\ & = \left|\frac{a_{n+1}}{a_{n}}\right| \\ & =\left|\frac{(4-3 i)^{n+1}}{((n+1)+1) !} \bigg/ \frac{(4-3 i)^{n}}{(n+1) !}\right| \\ & = \frac{|4-3 i|^{n+1}}{|4-3 i|^{n}} \frac{(n+1) !}{(n+2) !} \\ & = \frac{5}{n+2} \\ & = 0 \end{align*}\] By the Ratio Test we conclude that the series \(\sum_{n=1}^\infty a_n\) converges absolutely, and hence converges.

8.5.3 Exponential function and Euler’s Number

We have already encountered the exponential function in the Euler’s identity \[ e^{i\theta} = \cos(\theta) + i \sin (\theta) \,, \quad \, \theta \in \mathbb{R}\,. \] Using the Ratio Test for general series, we can give a precise definition for \(e^z\) with \(z \in \mathbb{C}\). We start by studying convergence of the exponential series.

Theorem 47: Exponential series

Let \(z \in \mathbb{C}\). The exponential series \[ \sum_{n=0}^{\infty} \frac{z^{n}}{n !} \] converges absolutely.

Proof

Set \[ a_{n}=\frac{z^{n}}{n !} \,. \] Then \[\begin{align*} L & = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_{n}}\right| \\ & = \lim_{n \to \infty} \left|\frac{z^{n+1}}{(n+1) !} \bigg/ \frac{z^{n}}{n !}\right| \\ & = \lim_{n \to \infty} \frac{|z|^{n+1}}{|z|^{n}} \frac{n !}{(n+1) !} \\ & = \lim_{n \to \infty} \frac{|z|}{n+1} \\ & = |z| \cdot \lim_{n \to \infty} \frac{1}{n+1} = 0 \,. \end{align*}\] Therefore the series converges absolutely by the Ratio Test in Theorem 45.

Definition 48: Exponential function

Define the exponential function \[ \exp \colon \mathbb{C}\to \mathbb{C}\,, \quad \exp (z) := \sum_{n=0}^{\infty} \frac{z^{n}}{n !} \] for \(z \in \mathbb{C}\). We denote \[ e^{z} := \exp(z) \,, \quad e:=e^{1} \,. \]

Remark 49

Using the definition of \(e^z\), one can show the usual properties of exponentials, such that \[ e^{x+y}=e^{x} e^{y} \,, \quad \frac{\mathrm{d}}{\mathrm{d} x} e^{x}=e^{x} \,. \]
We had defined \[ e := \lim_{n \to \infty } \left( 1 + \frac1n \right)^n \,. \] Using the binomial theorem one can prove that \[ \lim_{n \to \infty } \left( 1 + \frac1n \right)^n = \sum_{n=0}^{\infty} \frac{1}{n !} \,. \]

8.5.4 Conditional convergence

Some series do not converge absolutely, but still converge. Such series are said to converge conditionally.

Definition 50: Conditional convergence

Let \((a_n)\) be a sequence in \(\mathbb{C}\). We say that the series \[ \sum_{n=1}^{\infty} a_{n} \] converges conditionally if it converges, but it does not converge absolutely.

In practice conditional convergence means that the convergence of the series depends on the order in which we perform the summation. Changing the order of summation of a series is called rearrangement.

Definition 51: Rearrangement of a series

Let \((a_n)\) be a sequence in \(\mathbb{C}\). We define:

A permutation is a bijection \(\sigma \colon \mathbb{N}\to \mathbb{N}\).
A rearrangement of the series \(\sum_{n=1}^\infty a_n\) is a series \[ \sum_{n=1 }^\infty a_{\sigma(n)} \] for some permutation \(\sigma\).

If a series of complex numebers converges absolutely, then all its rearrangements converge to the same limit.

Theorem 52

Let \((a_n)\) be a sequence in \(\mathbb{C}\) such that \[ \sum_{n=1}^\infty |a_n| \] converges. For any permutation \(\sigma\) we have \[ \sum_{n=1}^\infty a_{\sigma(n)} = \sum_{n=1}^\infty a_n \, . \]

For a proof, see Theorem 3.55 in (Rudin 1976). A very surprising result is the following: If a series of real numbers converges conditionally, then the series can be rearranged to converge to any real number.

Theorem 53: Riemann rearrangement Theorem

Let \((a_n)\) be a real series such that \[ \sum_{n=1}^\infty a_n \] converges conditionally. Let \[ L \in \mathbb{R}\,\, \mbox{ or } \,\, L = \pm \infty \,. \] There exists a permutation \(\sigma\) such that the corresponding rearrangement \(\sum_{n=1}^\infty a_{\sigma(n)}\) converges conditionally to \(L\), that is, \[ \sum_{n=1}^\infty a_{\sigma(n)} = L \, . \]

For a proof, we refer the reader to Theorem 3.54 in (Rudin 1976).

8.5.5 Dirichlet and Alternating Series Tests

There are very few conditional convergence tests available. We present the Dirichlet Test and the Alternating Series Test.

Theorem 54: Dirichlet Test

Let \((c_n)\) be a sequence in \(\mathbb{C}\) and \((q_n)\) a sequence in \(\mathbb{R}\). Suppose that

\(q_n\) is decreasing,
\(q_n \to 0\),
\(q_n \geq 0\) for all \(n \in \mathbb{N}\).
Suppose there exists \(M > 0\) such that \[ \left| \sum_{n=1}^k c_n \right| \leq M \,, \quad \forall \, k \in \mathbb{N}\,. \]

Then the following series converges \[ \sum_{n=1}^\infty c_n q_n \]

Proof

Define the partial sums \[ s_k := \sum_{n=1}^k c_n \,, \quad \forall \, k \in \mathbb{N}\,. \] By assumption it holds \[ |s_k| \leq M \,, \quad \forall \, k \in \mathbb{N}\,. \tag{8.12}\] Note that \[\begin{align*} c_1 & = s_1 \\ c_2 & = s_2 - s_1 \\ & \ldots \ldots \\ c_n & = s_{n} - s_{n-1} \,. \end{align*}\] Therefore \[\begin{align*} \sum_{n=1}^k c_k q_k & = c_1 q_1 + c_2 q_2 + \ldots + c_k q_k \\ & = s_1 q_1 + (s_2 - s_1) q_2 + \ldots + (s_k - s_{k-1})q_k \\ & = s_1 (q_1 - q_2) + s_2 (q_2 - q_3) + \ldots + s_{k-1} (q_{k-1} - q_k) + s_k q_k \\ & = \left( \sum_{n=1}^{k-1} s_n (q_n - q_{n+1}) \right) + s_k q_k \end{align*}\] Since \((s_k)\) is bounded and \(q_k \to 0\), we conclude that \[ s_k q_k \to 0 \,. \] Further, notice that \[ q_n - q_{n+1} \geq 0 \,, \] since \(q_n\) is decreasing. Therefore, using (8.12), we get \[\begin{align*} |s_n| |q_n - q_{n+1}| & = |s_n| (q_n - q_{n+1}) \\ & \leq M (q_n - q_{n+1}) \end{align*}\] for all \(n \in \mathbb{N}\). Note that \[ \sum_{n=1}^{k-1} M (q_n - q_{n+1}) = M (q_1 - q_{k}) \,. \] Since \(q_k \to 0\) as \(k \to \infty\), we conclude that \[ \sum_{n=1}^{\infty} M (q_n - q_{n+1}) = M q_1 \,. \] Hence, by the Comparison Test for non-negative series, we infer that \[ \sum_{n=1}^\infty |s_n| |q_n - q_{n+1}| \,. \] In particular the series \[ \sum_{n=1}^\infty s_n (q_n - q_{n+1}) \] converges by the Absolute Convergence Test. Since we have shown \[ \sum_{n=1}^k c_k q_k = \left( \sum_{n=1}^{k-1} s_n (q_n - q_{n+1}) \right) + s_k q_k \] and \(s_k q_k \to 0\), we conclude that \[ \sum_{n=1}^\infty c_k q_k \] converges.

Example 55

Let \(\theta \in \mathbb{R}\), with \[ \theta \neq 2 k \pi \,, \quad \forall \, k \in \mathbb{Z}\,. \] Prove that the below series are conditionally convergent \[ \sum_{n=1}^\infty \frac{e^{i \theta n}}{n} \,, \quad \sum_{n=1}^{\infty} \frac{\cos(\theta n)}{n} \,, \quad \sum_{n=1}^{\infty} \frac{\sin(\theta n)}{n} \,. \]

Solution. Clearly the series \[ \sum_{n=1}^\infty \frac{e^{i \theta n}}{n} \] does not converge absolutely, since \[ \sum_{n=1}^\infty \left| \frac{e^{i \theta n}}{n} \right| = \sum_{n=1}^\infty \frac{1}{n} \] diverges, being the Harmonic Series. Let us prove convergence of \(\sum_{n=1}^\infty \frac{e^{i \theta n}}{n}\). Define the sequences \[ c_n := e^{i\theta n} \,, \quad q_n := \frac{1}{n} \,. \] We have that \(q_n\) is decreasing, \(q_n \to 0\) and \(q_n \geq 0\). Let us prove that there exists \(M>0\) such that \[ \left| \sum_{n=1}^k e^{i\theta n} \right| \leq M \,,\quad \forall \, k \in \mathbb{N}\,. \tag{8.13}\] Note that \[ 1 - e^{i\theta} \neq 0 \,, \] since \(\theta \neq 2k\pi\) for all \(k \in \mathbb{Z}\). Therefore we can use the Geometric Series (truncated) summation formula to get \[\begin{align*} \sum_{n=1}^k e^{i\theta n} & = \sum_{n=1}^k e^{i\theta n} \\ & = \frac{1- e^{i(k+1)\theta}}{1 - e^{i\theta}} - 1 \\ & = e^{i\theta} \, \frac{1- e^{ik\theta}}{1 - e^{i\theta}} \end{align*}\] Thus \[\begin{align*} \left| \sum_{n=1}^k e^{i\theta n} \right| & = \left| e^{i\theta} \, \frac{1- e^{ik\theta}}{1 - e^{i\theta}} \right| \\ & = \left| e^{i \theta} \right| \left|\frac{1- e^{ik\theta}}{1 - e^{i\theta}} \right| \\ & = \frac{ |1- e^{ik\theta}|}{|1 - e^{i\theta} |} \\ & \leq \frac{ |1| + |e^{ik\theta}|}{|1 - e^{i\theta} |} \\ & = \frac{ |1| }{|1 - e^{i\theta} |} \end{align*}\] where we used the triangle inequality. Since the right hand side does not depend on \(k\), we can set \[ M := \frac{ |1| }{|1 - e^{i\theta} |} \,, \] so that (8.13) holds. Therefore \[ \sum_{n=1}^\infty \frac{e^{i \theta n}}{n} \] converges by the Dirichlet Test. Recalling the Euler’s Identity \[ e^{i\theta} = \cos(\theta) + i \sin(\theta) \,, \] we obtain that also the series \[ \sum_{n=1}^\infty \frac{\cos(n \theta)}{n} \,, \quad \sum_{n=1}^\infty \frac{\sin(n \theta)}{n} \] converge.

As a corollary of the Dirichlet Test we obtain the Alternate Convergence Test.

Theorem 56: Alternate Convergence Test

Let \((q_n)\) be a sequence in \(\mathbb{R}\) such that

\(q_n\) is decreasing,
\(q_n \to 0\),
\(q_n \geq 0\) for all \(n \in \mathbb{N}\).

The following series converges \[ \sum_{n=1}^\infty (-1)^n q_n \]

Proof

Define the sequence \[ c_n := (-1)^n \,. \] Then \[ \sum_{n=1}^k c_n = \begin{cases} 0 & \,\, \mbox{ if } \,\, k \, \mbox{ even} \\ -1 & \,\, \mbox{ if } \,\, k \, \mbox{ odd} \\ \end{cases} \] Hence \[ \left| \sum_{n=1}^k c_n \right| \leq 1 \,, \quad \forall \, k \in \mathbb{N}\,. \] By the Dirichlet Test we have convergence of \[ \sum_{n=1}^\infty c_n q_n = \sum_{n=1}^\infty (-1)^n q_n \,. \]

Example 57

Prove that the below series converges conditionally \[ \sum_{n=1}^{\infty}(-1)^{n} \frac{1}{n} \,. \]

Solution. The series does not converge absolutely, since \[ \sum_{n=1}^{\infty} \left| (-1)^{n} \frac{1}{n} \right| = \sum_{n=1}^\infty \frac1n \] diverges, being the Harmonic Series. We need to show convergence of \[ \sum_{n=1}^{\infty}(-1)^{n} \frac{1}{n} \,. \] To this end, set \[ q_n := \frac1n \,. \] Clearly \(q_n \geq 0\), \(q_n \to 0\) and \(q_n\) is decreasing. Hence the series converges by the Alternating Series Test.

8.5.6 Abel’s Test

Another conditional convergence test is the Abel Test. This looks similar to the Dirichlet Test, however notice that the Abel Test only deals with real sequences.

Theorem 58: Abel’s Test

Let \((a_n)\) and \((q_n)\) be sequences in \(\mathbb{R}\). Suppose that

\(q_n\) is monotone,
\(q_n\) is bounded,
The series below converges \[ \sum_{n=1}^\infty a_n \,. \]

Then the following series converges \[ \sum_{n=1}^\infty a_n q_n \]

The proof is similar to the one of the Dirichlet Test. We decided to omit it.

Example 59

Prove that the series below converges conditionally \[ \sum_{n=1}^\infty \ \frac{(-1)^n}{n} \left( 1 + \frac{1}{n} \right)^n \,. \]

Solution. Set \[ a_n:= \frac{(-1)^n}{n}\,, \quad q_n := \left( 1 + \frac{1}{n} \right)^n \,. \] We have seen that \(q_n\) is monotone increasing and bounded. Moreover the series \(\sum_{n=1}^\infty a_n\) converges by the Alternating Series Test, as seen in Example 57. Hence the series \(\sum_{n=1}^\infty a_n q_n\) converges by the Abel Test.

However the series in question does not converge absolutely. Indeed, \[ \left|\frac{(-1)^n}{n} \left( 1 + \frac{1}{n} \right)^n \right| = \frac{1}{n} q_n \geq \frac{1}{n} q_1 \,, \] since \((q_n)\) is increasing. As the series \[ \sum_{n=1}^\infty \frac{1}{n} q_1 \] diverges, by the Comparison Test we conclude that also \[ \sum_{n=1}^\infty \left|\frac{(-1)^n}{n} \left( 1 + \frac{1}{n} \right)^n \right| \] diverges, proving that the series in the example converges conditionally.