7 Series

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.

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

Question. Prove that \[ \sum_{n=1}^{\infty} \frac{1}{n(n+1)} = 1 \,. \]

Solution. The idea to prove convergence is to split the general term into the sum of two fraction: \[\begin{align*} \frac{1}{n(n+1)} & = \frac{A}{n} + \frac{B}{n(n+1)} \\ & = \frac{A(n+1) + Bn}{n(n+1)} \\ & = \frac{(A+B)n+A}{n(n+1)} \,. \end{align*}\] In order for the LHS and RHS to be the same, we need to impose \[ (A+B)n+A = 1 \,, \] which holds if and only if \[ A+ B = 1 , \,\, A = 1 \quad \implies \quad A =1 ,\,\, B = - 1\,. \] Therefore, we conclude that \[ \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1} \,. \] We can now compute the partial sums \(s_{k}\) as follows: \[\begin{align*} s_{k} & =\sum_{n=1}^{k} \frac{1}{n(n+1)} \\ & =\sum_{n=1}^{k}\left(\frac{1}{n}-\frac{1}{n+1}\right) \\ & =\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{k}-\frac{1}{k+1} \\ & =1-\frac{1}{k+1} . \end{align*}\] Therefore, \[ \lim_{k \rightarrow \infty} s_{k}=\lim _{k \rightarrow \infty}\left(1-\frac{1}{k+1}\right)=1 \,, \] which means that 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 series, since we can fold the entire partial sum together, in such a way that only two terms remain.

Example 5

Question. Prove that the following series diverges \[ \sum_{n=1}^{\infty}(-1)^{n} \,. \]

Solution. 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 \(\sum (-1)^{n}\) diverges.

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

Example 7

Consider the series \[ \sum_{n=1}^\infty (-1)^n \,. \tag{7.1}\] 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 (7.1) diverges by Theorem 6.

Example 8

Question. Discuss converge/divergence for the following series \[ \sum_{n=1}^{\infty} \frac{n}{5 n+11} \,. \]

Solution. We have \[ a_{n}:=\frac{n}{5 n+11}=\frac{1}{5 + \dfrac{11}{n} } \longrightarrow \frac{1}{5} \neq 0 \,. \] Hence, the series \(\sum a_n\) diverges.

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

Question. Discuss convergence/divergence for the following series \[ \sum_{n=1}^\infty a_n \,, \quad a_n := \frac{1}{ \sqrt{n+1} + \sqrt{n} } \,. \]

Solution. By the Algebra of Limits we have \[ \lim_{n \to \infty} a_n = 0 \,. \] Therefore, we cannot conclude anything yet: The series might converge or diverge. Let us compute the partial sums: \[\begin{align*} s_k & = \sum_{n=1}^k \frac{1}{ \sqrt{k+1} + \sqrt{k} } \\ & = \sum_{n=1}^k \frac{1}{ \sqrt{k+1} + \sqrt{k} } \, \cdot \, \frac{\sqrt{k+1} - \sqrt{k}}{ \sqrt{k+1} - \sqrt{k} } \\ & = \sum_{n=1}^k \sqrt{k+1} - \sqrt{k} \\ & = \sqrt{2} - \sqrt{1} + \sqrt{3} - \sqrt{2} + \ldots + \sqrt{k+1} - \sqrt{k} \\ & = \sqrt{k+1} - 1 \,. \end{align*}\] We have shown that the partial sums are \[ s_k = \sum_{n=1}^k a_n = \sqrt{k+1} - 1 \,. \] Therefore \((s_k)\) is divergent, and so the series \(\sum a_n\) is divergent.

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

Question. Prove that \[ \sum_{n=7}^{\infty} \frac{1}{n(n+1)} = \frac17 \,. \]

Solution. 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. 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 \[ \sum_{n=7}^{\infty} \frac{1}{n(n+1)} = \lim_{k \rightarrow \infty} s_{k} = \frac{1}{7} \,. \]

7.1 Geometric series

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

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

Theorem 13: Geometric Series Test

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

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

\[ \sum_{n=0}^{\infty} x^{n}=\frac{1}{1-x} \,. \tag{7.2}\]

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

Example 14

Question. Discuss convergence/divergence of the following series. If the series converges, compute the limit. \[\begin{align*} & \sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{n}\,, & \quad \sum_{n=0}^{\infty}\left(-\frac{3}{2}\right)^{n} \,, \\ & \sum_{n=0}^{\infty}\left(\frac{-3}{4}\right)^{n}\,, & \quad \sum_{n=0}^{\infty}(-1)^{n} \,. \end{align*}\]

Solution.

Since \(\left|\frac{1}{2}\right|<1\), by the GST we have \[ \sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{n}=\frac{1}{1-\dfrac{1}{2}}=2 \]
Since \(\left|\frac{-3}{2}\right|=\frac{3}{2}>1\), by the GST the series \[ \sum_{n=0}^{\infty}\left(-\frac{3}{2}\right)^{n} \] diverges.
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 \[ \sum_{n=0}^{\infty}(-1)^{n} \] diverges.

Remark 15

If the sum of a Geometric Sries does not start at \(n=0\), we need to tweak the summation formula at (7.2). 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

Question. Prove that \[ \sum_{n=1}^{\infty}\left(\frac{1}{2}\right)^{n} = 1 \,. \]

Solution. We have that \[\begin{align*} \sum_{n=1}^{\infty}\left(\frac{1}{2}\right)^{n} & = \sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{n} - 1 \\ & = \frac{1}{1- \dfrac{1}{2}} - 1 = 1\,. \end{align*}\]

Example 17

Question. Discuss convergence/divergence of the following series. If the series converges, compute the limit. \[ \sum_{n=0}^{\infty} \frac{1}{(1+i)^{n}} \,, \quad \sum_{n=0}^{\infty}\left(\frac{1-5 i}{3+3 i}\right)^{n} \,, \quad \sum_{n=0}^{\infty}\left(\frac{2+i}{3-2 i}\right)^{n} \,. \]

Solution.

We have \[ \frac{1}{(1+i)^{n}}=\left(\frac{1}{1+i}\right)^{n} \] 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 \[ \sum_{n=0}^{\infty} \frac{1}{(1+i)^{n}} = \frac{1}{1-\dfrac{1}{1+i}} = 1-i \,. \]
Since \[\begin{align*} \left|\frac{1-5 i}{3+3 i}\right| & = \frac{|1-5 i|}{|3+3 i|} \\ & =\frac{\sqrt{(1)^{2}+(-5)^{2}}}{3\sqrt{1^{2}+1^{2}}} \\ & = \frac{\sqrt{26}}{3\sqrt{2}} \\ & =\frac{\sqrt{13}}{3} > 1 \,, \end{align*}\] the series diverges by the Geometric Series Test.
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 \[\begin{align*} \sum_{n=0}^{\infty}\left(\frac{2+i}{3-2 i}\right)^{n} & = \frac{1}{1-\dfrac{2+i}{3-2 i}} \\ & = \frac{1}{\dfrac{3-2 i-(2+i)}{3-2 i}} \\ & = \frac{3-2 i}{1-3 i} \\ & = \frac{3-2 i}{1-3 i} \, \frac{1+3 i}{1+3 i} \\ & = \frac{3-2 i+9 i-6 i^{2}}{1-9 i^{2}} \\ & = \frac{9}{10} + \frac{7}{10} i \end{align*}\]

7.2 Algebra of Limits for Series

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 \[ \sum_{n=1}^{\infty} c \cdot a_{n}= c \cdot a \,. \]

Example 19

Question. Prove that \[ \sum_{n=0}^{\infty}\left(2\left(\frac{1}{3}\right)^{n}+\left(\frac{2}{3}\right)^{n}\right) = 6 \,. \]

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

7.3 Non-negative series

Definition 20: 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 \[ a_n \geq 0 \,, \quad \forall \, n \in \mathbb{N}\,. \]

Lemma 21

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.

We present 4 test for the convergence of non-negative series:

Cauchy Condensation Test
Comparison Test
Limit Comparison Test
Ratio Test (positive series only)

Theorem 22: 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 following series converges \[ \sum_{n=1}^\infty a_n \,. \]
The following series converges \[ \sum_{n=0}^\infty 2^n a_{2^n} = a_1 + 2a_2 + 8a_8 + 16 a_{16} + \ldots \]

Theorem 23: 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 22, 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 if and only if \[ |x| < 1 \,, \] which is equivalent to \[\begin{align*} 2^{1-p} < 1 = 2^0 \quad & \iff \quad 1 - p < 0 \\ & \iff \quad p > 1 \,. \end{align*}\] Therefore \[ \sum_{n=1}^\infty \frac{1}{n^p} \] converges if and only if \(p>1\), ending the proof.

Theorem 24

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 23 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.

Theorem 25: 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: \[\begin{align*} \sum_{n=1}^{\infty} b_{n} & \,\, \text { converges } \Longrightarrow \quad \sum_{n=1}^{\infty} a_{n} \,\, \text { converges, } \\ \sum_{n=1}^{\infty} a_{n} & \,\, \text { diverges } \Longrightarrow \quad \sum_{n=1}^{\infty} b_{n} \,\, \text { diverges. } \end{align*}\]

Example 26

Question. Discuss convergence/divergence of the following series: \[ \sum_{n=1}^{\infty} \frac{1}{n^{2}+3 n-1} \,, \tag{7.3}\] \[ \sum_{n=0}^{\infty} \frac{3^{n}+6 n+\dfrac{1}{n+1}}{2^{n}} \,. \tag{7.4}\]

Solution.

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 23 the \(p\)-series \[ \sum_{n=1}^{\infty} \frac{1}{n^{2}} \] converges. Therefore also the series at (7.3) converges by the Comparison Test in Theorem 25.
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 (7.4) diverges.

Theorem 27: 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 \[\begin{align*} \sum_{n=1}^{\infty} b_{n} & \,\, \text { converges } \Longrightarrow \quad \sum_{n=1}^{\infty} a_{n} \,\, \text { converges, } \\ \sum_{n=1}^{\infty} a_{n} & \,\, \text { diverges } \Longrightarrow \quad \sum_{n=1}^{\infty} b_{n} \,\, \text { diverges. } \end{align*}\]

Example 28

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

Solution. 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 \[ \sum_{n=1}^{\infty} \frac{2 n^{3}+5 n+1}{7 n^{6}+2 n+5} \] converges, by the Limit Comparison Test.

Example 29

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

Solution. Since \(\sin(n)\) is bounded, 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 \[\begin{align*} L & := \frac{a_n}{b_n} = \lim_{n \rightarrow \infty} \frac{n+\cos (n)}{n^{2}} \bigg/ \frac{1}{n} \\ & = \lim_{n \rightarrow \infty} \frac{n^{2}+n \cos (n)}{n^{2}} \\ & = \lim_{n \rightarrow \infty} \left(1+\frac{\cos (n)}{n}\right) \end{align*}\] Note that \[ -1 \leq \cos (n) \leq 1 \quad \implies \quad -\frac{1}{n} \leq \frac{\cos (n)}{n} \leq \frac{1}{n} \,. \] 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 \(\sum_{n=1}^{\infty} \frac{1}{n}\) diverges. Since \(L = 1 > 0\), the series \[ \sum_{n=1}^{\infty} \frac{n+\cos (n)}{n^{2}} \,. \] diverges by the Limit Comparison Test.

Example 30

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

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 \[ L := \lim_{n \to \infty} \frac{b_n}{c_n} = \lim_{n \to \infty} n^2 \, \left(1 - \cos \left( \frac{1}{n} \right) \right) = \frac{1}{2} \,. \] 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.

Example 31

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

Solution. Since \[ \sin (n) \geq-1 \,, \] the above is a non-negative series. As \(\sin(n)\) is bounded, the series behaves similarly 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 25.

Theorem 32: 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.

Example 33

Question. Discuss convergence/divergence of the following series \[ \sum_{n=1}^{\infty} a_n \,, \quad a_n = \frac{(n !)^{2}}{(2 n) !} \,. \]

Solution. We compute \[\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_{n}} & = \lim_{n \to \infty} \frac{((n+1) !)^{2}}{(2(n+1)) !} \bigg/ \frac{(n !)^{2}}{(2 n) !} \\ & = \lim_{n \to \infty} \frac{(n+1)^{2}}{(2 n+2)(2 n+1)} \\ & = \lim_{n \to \infty} \frac{\left(1+\dfrac{1}{n}\right)^{2}}{\left(2+\dfrac{2}{n}\right)\left(2+\dfrac{1}{n}\right)} = \frac14 \,. \end{align*}\] Since \(L = 1/4 <1\), by the Ratio Test we conclude that \(\sum a_n\) converges.

Example 34

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

Solution. Set \(a_{n}=\log n / n^{2}\). By the Cauchy Condensation Test, we know that \(\sum a_n\) converges if and only if \(\sum 2^{n} a_{2^{n}}\) converges. We have: \[\begin{align*} \sum_{n=0}^{\infty} 2^{n} a_{2^{n}} & =\sum_{n=0}^{\infty} 2^{n} \frac{\log (2^{n})}{\left(2^{n}\right)^{2}} \\ & = \log (2) \, \sum_{n=0}^{\infty} \frac{n }{2^{n}} \\ & = \log (2) \, \sum_{n=0}^{\infty} b_n \,, \qquad b_n := \frac{n }{2^{n}} \,. \end{align*}\] Apply the Ratio Test to the series \(\sum b_n\) \[ \frac{b_{n+1}}{b_n} = \frac{n+1}{2^{n+1}} \bigg/ \frac{n}{2^n} = \frac{n+1}{2n} \longrightarrow \frac12 < 1 \,. \] Therefore, \(\sum b_n\) converges by the Ratio Test, so that also \(\sum 2^n a_{2^n}\) converges. We conclude that \(\sum a_n\) converges by the Cauchy Condensation Test.

7.4 General series

Definition 35: 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 following non-negative series converges \[ \sum_{n=1}^{\infty}\left|a_{n}\right| \,. \]

Theorem 36: 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.

Example 37

Question. Discuss absolute convergence of the series \[ \sum_{n=1}^{\infty}(-1)^{n} \frac{1}{n} \,. \]

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

Example 38

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

Solution. We have \[ |a_n| = \frac{\left|n^{2}-5 n+2\right|}{n^{4}} = \frac{n^{2}+5 n+2}{n^{4}} \,, \] for \(n\) sufficiently large (e.g. \(n \geq 10\)). 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 \(\sum 1/n^2\) 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). This shows \(\sum |a_n|\) converges, which means that \(\sum a_n\) converges absolutely. In particular, \(\sum a_n\) converges by the Absolute Convergence Test.

Theorem 39: 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.

Example 40

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

Solution. We have \[\begin{align*} L & := \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \\ & =\lim_{n \to \infty} \left|\frac{(4-3 i)^{n+1}}{((n+1)+1) !} \bigg/ \frac{(4-3 i)^{n}}{(n+1) !}\right| \\ & = \lim_{n \to \infty} \frac{5}{n+2} = 0 \,.\\ \end{align*}\] As \(L=0<1\), we conclude that \(\sum a_n\) converges absolutely, by the Ratio Test. Hence, \(\sum a_n\) converges by the Absolute Convergence Test.

Theorem 41: Exponential series

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

Proof

Set \(a_{n}=z^n / n!\). Then \[\begin{align*} L & = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_{n}}\right| \\ & = \lim_{n \to \infty} \left|\frac{z^{n+1}}{(n+1) !} \bigg/ \frac{z^{n}}{n !}\right| \\ & = \lim_{n \to \infty} \frac{|z|}{n+1} = 0 \,.\\ \end{align*}\] Therefore the series converges absolutely by the Ratio Test in Theorem 40.

7.5 Conditional convergence

Definition 42: 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.

Definition 43: Rearrangement of a series

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

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 \(\sigma\).

Theorem 44

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

Theorem 45: Riemann rearrangement Theorem

Let \((a_n)\) be a real sequence such that the series \[ \sum_{n=1}^\infty a_n \] 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, \[ \sum_{n=1}^\infty a_{\sigma(n)} = L \, . \]

Theorem 46: 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 \[ \sum_{n=1}^\infty c_n q_n \,. \]

Example 47

Question. 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.

Recalling the Euler’s Identity \[ e^{i\theta} = \cos(\theta) + i \sin(\theta) \,, \] we obtain that \[ \sum_{n=1}^\infty \frac{e^{i \theta n}}{n} = \sum_{n=1}^\infty \frac{\cos(n \theta)}{n} + i \sum_{n=1}^\infty \frac{\sin(n \theta)}{n} \,. \] Therefore, the series \(\sum e^{i \theta n}/n\) converge conditionally if and only if \(\sum \cos( \theta n)/n\) and \(\sum \sin( \theta n)/n\) converge conditionally. It is then sufficient to study \(\sum e^{i \theta n}/n\).
The series \(\sum e^{i \theta n} /n\) 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.
Set \(c_n = e^{i\theta n}\), \(q_n = 1/n\), so that \[ \sum_{n=1}^\infty \frac{e^{i \theta n}}{n} = \sum_{n=1}^\infty c_nq_n \,. \] 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 \[ \left| \sum_{n=1}^k e^{i\theta n} \right| \leq M \,,\quad \forall \, k \in \mathbb{N}\,. \tag{7.5}\] 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 \[\begin{align*} \sum_{n=1}^k e^{i\theta n} & = \sum_{n=1}^k (e^{i\theta})^n \\ & = \frac{1- e^{i(k+1)\theta}}{1 - e^{i\theta}} - 1 \\ & = e^{i\theta} \, \frac{1- e^{ik\theta}}{1 - e^{i\theta}} \end{align*}\] Taking the modulus \[\begin{align*} \left| \sum_{n=1}^k e^{i\theta n} \right| & = \left| e^{i\theta} \, \frac{1- e^{ik\theta}}{1 - e^{i\theta}} \right| = \left| e^{i \theta} \right| \left|\frac{1- e^{ik\theta}}{1 - e^{i\theta}} \right| \\ & = \frac{ |1- e^{ik\theta}|}{|1 - e^{i\theta} |} \leq \frac{ |1| + |e^{ik\theta}|}{|1 - e^{i\theta} |} = \frac{ 2 }{|1 - e^{i\theta} |} \,, \end{align*}\] where we used the triangle inequality. Since the RHS does not depend on \(k\), we can set \[ M = \frac{ 2 }{|1 - e^{i\theta} |} \,, \] so that (7.5) holds. Therefore, \(\sum e^{i \theta n}/n\) converges by the Dirichlet Test.
We have shown that \(\sum e^{i \theta n}/n\) converges, but not absolutely. Hence, it converges conditionally.

Theorem 48: 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 \[ \sum_{n=1}^\infty (-1)^n q_n \]

Example 49

Question. Prove that the 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. Set \(q_n = 1/n\), so that \[ \sum_{n=1}^{\infty}(-1)^{n} \frac{1}{n} = \sum_{n=1}^\infty (-1)^n q_n \,. \] Clearly, \(q_n \geq 0\), \(q_n \to 0\) and \(q_n\) is decreasing. Hence, the series converges by the Alternating Series Test. Thus, the series converges conditionally.

Theorem 50: Abel’s Test

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

\(q_n\) is monotone and bounded,
The series \(\sum a_n\) converges.

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

Example 51

Question. Prove that the series 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 (recall that \(q_n \to \varepsilon\)). Moreover, the series \(\sum_{n=1}^\infty a_n\) converges by the Alternating Series Test, as seen in Example 49. 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, \[ \left|\frac{(-1)^n}{n} \left( 1 + \frac{1}{n} \right)^n \right| = \frac{1}{n} q_n \geq \frac{1}{n} q_1 = \frac{2}{n}\,, \] since \((q_n)\) is increasing. As the series \(\sum 2/n\) 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. Therefore, the series in the example converges conditionally.

Good Luck with the Exam!