Absolute convergence. The series \(\sum_{n=1}^{\infty}a_n\) is said to converge absolutely if the series \[\sum_{n=1}^{\infty}|a_n|<\infty\] converges.
Theorem. If \(\sum_{n=1}^{\infty}|a_n|<\infty\), then \(\left|\sum_{n=1}^{\infty}a_n\right|<\infty\).
Proof. The claim follows from the Cauchy Criterion, since \[\left|\sum_{k=m}^{n}a_k\right| \leq \sum_{k=m}^{n}|a_k|\] and we are done. $$\tag*{$\blacksquare$}$$
Conditional convergence. If the series \(\sum_{n=1}^{\infty}a_n\) converges but \[\sum_{n=1}^{\infty}|a_n|=\infty\] diverges then we say that \(\sum_{n=1}^{\infty}a_n\) converges conditionally.
Example 1. For series with positive terms, absolute convergence is the same as convergence.
Example 2. \(\sum_{n=1}^{\infty}(-1)^n\frac{1}{n^2}\) converges absolutely, since \(\sum_{n=1}^{\infty}\frac{1}{n^2}<\infty\).
Anharmonic series. The series \[\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\] converges conditionally.
It is easy to see that \[\sum_{n=1}^{\infty}\left|\frac{(-1)^n}{n}\right|=\sum_{n=1}^{\infty}\frac{1}{n}=\infty.\] To prove \(\big|\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\big|<\infty\) we will show a more general result.
Abel summation formula. Given two sequences \((a_n)_{n \in \mathbb{N}}\) and \((b_n)_{n \in \mathbb{N}}\) set \[A_n=\sum_{k=0}^{n}a_k \quad \text{ for } \quad n \geq 0, \quad \text{ and } \quad A_{-1}=0.\]
Then if \(0 \leq p \leq q\) one has \[{\color{blue}\sum_{n=p}^{q}a_nb_n=\sum_{n=p}^{q-1}A_n \left(b_n-b_{n+1}\right)+A_qb_q-A_{p-1}b_p}.\]
Proof: Note that \[\begin{aligned} \sum_{n=p}^{q}a_nb_n&= \sum_{n=p}^q\underbrace{(A_n-A_{n-1})}_{{\color{red}a_n}}b_n= \sum_{n=p}^{q}A_nb_n-\sum_{n=p}^q A_{n-1}b_n \\ &= \sum_{n=p}^{q}A_nb_n-\sum_{n=p-1}^{q-1} A_{n}b_{n+1} \\&=\sum_{n=p}^{q-1}A_n \left(b_n-b_{n+1}\right)+A_qb_q-A_{p-1}b_p. \end{aligned}\] The proof follows. $$\tag*{$\blacksquare$}$$
Dirichlet’s test. Suppose that
The partial sums \(A_n=\sum_{k=1}^na_k\) of \((a_n)_{n\in \mathbb{N}}\) form a bounded sequence.
\(b_0 \geq b_1 \geq b_2 \geq b_3 \geq \ldots\),
\(\lim_{n \to \infty}b_n=0.\)
Then \(\sum_{n=1}^{\infty}a_nb_n\) converges.
Proof. Choose \(M \geq 0\) so that \(|A_n| \leq M\) for all \(n \in \mathbb{N}\). Given \(\varepsilon>0\) there is \(N_{\varepsilon} \in \mathbb{N}\) so that \[b_{N_{\varepsilon}}<\frac{\varepsilon}{2M},\] since \(\lim_{n \to \infty}b_n=0\).
For \(N_{\varepsilon} \leq p \leq q\), by the summation by parts formula, one has \[\begin{aligned} \left|\sum_{n=p}^q a_nb_n\right|&=\left|\sum_{n=p}^{q-1}A_n \left(b_n-b_{n+1}\right)+A_qb_q-A_{p-1}b_p\right| \\&\leq {\color{red}\left|\sum_{n=p}^{q-1}A_n \left(b_n-b_{n+1}\right)\right|}+{\color{blue}|A_qb_q|}+{\color{purple}|A_{p-1}b_p|} \\&\leq M {\color{red}\sum_{n=p}^{q-1}\left| \left(b_n-b_{n+1}\right)\right|}+M{\color{blue}b_q}+M{\color{purple}b_p} \leq 2Mb_p \leq 2Mb_{N_{\varepsilon}}<\varepsilon. \end{aligned}\] since \[\begin{aligned} b_p&-b_q={\color{red}\sum_{n=p}^{q-1}\left| \left(b_n-b_{n+1}\right)\right| }=\sum_{n=p}^{q-1} \left(b_n-b_{n+1}\right) \\ &=(b_p-{\color{red}b_{p+1}})+({\color{red}b_{p+1}}-{\color{blue}b_{p+2}})+({\color{blue}b_{p+2}}-{\color{purple}b_{p+3}})+\ldots+b_{q-1}-b_q.\end{aligned}\qquad\blacksquare\]
We now show that \(\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\) converges. Let \[{\color{red}a_n=(-1)^n,} \quad \text{ and }\quad {\color{blue}b_n=\frac{1}{n}}\] in the previous theorem. We see that \[\left|\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\right|=\left|\sum_{n=1}^{\infty}a_nb_n\right|<\infty\] since \[|A_n|=\left|\sum_{k=1}^n(-1)^k\right| \leq 1.\]
A more general result can be proved:
Alternating Series Test. Let \((a_n)_{n \in \mathbb{N}}\) be such that
\(a_1 \geq a_2 \geq \ldots \geq a_n \geq \ldots\),
\(\lim_{n \to \infty}a_n=0\).
Then the alternating series \(\sum_{n=1}^{\infty}(-1)^n a_n\) converges.
Proof. We apply the previous theorem.
Exercise. Determine if the series \(\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt{n^2+1}}\) converges and converges absolutely.
Solution. Let \(a_n=\frac{1}{\sqrt{n^2+1}}\).
We have \[a_n \geq \frac{1}{\sqrt{4n^2}}=\frac{1}{2n},\] so the series does not converges absolutely, since \(\sum_{n=1}^{\infty}\frac{1}{n}\) diverges.
On the other hand, we have \[a_n \geq a_{n+1} \quad \text{ and }\quad \lim_{n \to \infty}a_n=0,\] so the assumptions of the previous theorem are satisfied. Hence \(\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt{n^2+1}}\) converges conditionally.$$\tag*{$\blacksquare$}$$
Definition. Given \(\sum_{n=0}^\infty a_n\) and \(\sum_{n=0}^\infty b_n\) we set \[{\color{blue}c_n=\sum_{k=0}^na_kb_{n-k}=a_0b_n+a_1b_{n-1}+\ldots+a_{n-1}b_1+a_nb_0}\] and call \(\sum_{n=0}^\infty c_n\) the product of the two given series
\[\left(\sum_{n=0}^\infty a_n\right)\left(\sum_{n=0}^\infty b_n\right)=\sum_{n=0}^\infty c_n.\]
Theorem. Suppose that
\(\sum_{n=0}^\infty |a_n|<\infty\),
\(\sum_{n=0}^\infty a_n=A\),
\(\sum_{n=0}^\infty b_n=B\),
\(c_n=\sum_{k=0}^n a_kb_{n-k}\).
Then \[\sum_{n=0}^{\infty}c_n=AB.\]
Proof. Let \[A_n=\sum_{k=0}^na_k, \ \ B_n=\sum_{k=0}^n b_k, \ \ C_n=\sum_{k=0}^n c_k,\] \[\beta_n=B_n-B.\] Then \[\begin{aligned} C_n&=a_0b_0+({\color{red}a_0b_1}+{\color{blue}a_1b_0})+\ldots+({\color{red}a_0b_n}+{\color{blue}a_1b_{n-1}}+\ldots+a_nb_0) \\&=a_0B_n+a_1 B_{n-1}+\ldots+a_nB_0 \\&=a_0(B+\beta_n)+a_1(B+\beta_{n-1})+\ldots+a_n(B+\beta_0). \end{aligned}\]
Thus \[C_n=A_nB+\underbrace{a_0\beta_n+a_1\beta_{n-1}+\ldots+a_n\beta_0}_{{\color{red}\gamma_n}}.\]
We will show that \(C_n \ _{\overrightarrow{n \to \infty}}\ AB\). Since \(A_nB \ _{\overrightarrow{n \to \infty}}\ AB\) it suffices to prove that \(\gamma_n \ _{\overrightarrow{n \to \infty}}\ 0\).
Set \(\alpha=\sum_{n=0}^{\infty}|a_n|\). Let \(\varepsilon>0\) be given. By (b) \(\beta_n \ _{\overrightarrow{n \to \infty}}\ 0\), thus we find \(N \in \mathbb{N}\) such that \[{\color{red}|\beta_n|<\varepsilon \quad \text{ for }\quad n \geq N},\] then \[\begin{aligned} |\gamma_n| &\leq |\beta_0 a_n+\ldots+\beta_N a_{n-N}|+|\beta_{N+1}a_{n-N+1}+\ldots+\beta_n a_0| \\& \leq|\beta_0 a_n+\ldots+\beta_N a_{n-N}|+\varepsilon \alpha. \end{aligned}\]
We keep \(N \in \mathbb{N}\) fixed and letting \(n \to \infty\) we obtain \[\limsup_{n \to \infty}|\gamma_n| \leq \varepsilon \alpha\] since \(a_k \ _{\overrightarrow{k \to \infty}}\ 0\). But \(\varepsilon>0\) is arbitrary we get \[\liminf_{n \to \infty}|\gamma_n|=\limsup_{n \to \infty}|\gamma_n|=0 =\lim_{n \to \infty}|\gamma_n|.\tag*{$\blacksquare$}\]
Remark. If \(\sum_{n=0}^\infty a_n=A\), \(\sum_{n=0}^{\infty}b_n=B\), \(\sum_{n=0}^\infty c_n=C\), and \[{\color{blue}c_n=\sum_{k=0}^na_kb_{n-k}=a_0b_n+a_1b_{n-1}+\ldots+a_{n-1}b_1+a_nb_0}\] then \(C=AB\).
Rearrangement. Let \((k_n)_{n \in \mathbb{N}}\) be a sequence in which every positive integer appears once and only once. Setting \[{\color{blue}a_n'=a_{k_n}}\] we say that \(\sum_{n=1}^{\infty}a_n'\) is rearrangement of \(\sum_{n=1}^{\infty}a_n\).
Example.
Consider the convergent series \[S=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}=1-\frac{1}{2}+\frac{1}{3}\underbrace{-\frac{1}{4}+\frac{1}{5}}_{{\color{blue} <0}}\underbrace{-\frac{1}{6}+\frac{1}{7}}_{{\color{blue}<0}}-\ldots\]
Consider also a rearrangement \(S'\) of \(S\) given by: \[\begin{aligned} S'&=1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\ldots+\\ &=\sum_{k=1}^{\infty}\bigg(\frac{1}{4k-3}+\frac{1}{4k-1}-\frac{1}{2k}\bigg) \end{aligned}\]
Observe that \(S<1-\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\) and \[\frac{1}{4k-3}+\frac{1}{4k-1}-\frac{1}{2k}>0 \quad \text{ for all } \quad k \in \mathbb{N}.\]
If \(S_n'\) is the partial sum of \(S'\) then \[S_3'<S_6'<S_9'<\ldots\] hence \(\limsup_{n \to \infty}S_n'>S_3'=\frac{5}{6}\).
Thus \(S'\) does not converge to \(S<\frac{5}{6}\).
Theorem. Let \(\sum_{n=1}^{\infty}a_n\) be a series that converges conditionally. Suppose that \[-\infty \leq {\color{red}\alpha} \leq {\color{blue}\beta} \leq +\infty.\]
Then there exists a reaarangement \(\sum_{n=0}^{\infty}a_n'\) with partial sums \(s_n'\) so that \[{\color{red}\liminf_{n \to \infty}s_n'=\alpha}, \quad \text{ and } \quad {\color{blue}\limsup_{n \to \infty}s'_n=\beta}.\]
Proof. Let \[p_n=\frac{|a_n|+a_n}{2} \geq 0, \quad \text{ and } \quad q_n=\frac{|a_n|-a_n}{2} \geq 0.\] Then \[p_n-q_n=a_n \quad \text{ and }\quad p_n+q_n=|a_n|.\]
Claim. The series \(\sum_{n=1}^{\infty}p_n\) and \(\sum_{n=1}^{\infty}q_n\) both diverge.
Indeed, if both were convergent then \[+\infty=\sum_{n=1}^{\infty}|a_n|=\sum_{n=1}^{\infty}(p_n+q_n)<+\infty,\] contradiction.
Since \[\sum_{n=1}^Na_n=\sum_{n=1}^{N}(p_n-q_n)=\sum_{n=1}^{N}p_n-\sum_{n=1}^N q_n,\] then divergence of \(\sum_{n=1}^{\infty}p_n\) and convergence of \(\sum_{n=1}^{\infty}q_n\) (or vice versa) implies divergence of \(\sum_{n=1}a_n\), contradiction.
Let \(P_1,P_2,\ldots\) denote the nonnegative terms of \(\sum_{n=1}^{\infty}a_n\) in the order in which they occur.
Let \(Q_1,Q_2,\ldots\) be the absolute values of the negative terms of \(\sum_{n=1}^{\infty}a_n\) also in their original order.
The series \(\sum_{n=1}^{\infty}P_n\) and \(\sum_{n=1}^{\infty}Q_n\) differ from \(\sum_{n=1}^{\infty}p_n\) and \(\sum_{n=1}^{\infty}q_n\) also by zero terms and therefore thay also diverge.
Claim. We shall construct \((m_n)_{n \in \mathbb{N}}\) and \((k_n)_{n \in \mathbb{N}}\) such that the series \[\begin{aligned} &{\color{red}P_1+\ldots+P_{m_1}}{\color{blue}-Q_1-\ldots-Q_{k_1}}\\ &\qquad {\color{red}+P_{m+1}+\ldots+P_{m_2}}{\color{blue}-Q_{k_1+1}-\ldots-Q_{k_2}}+\ldots \end{aligned}\] which is a rearrangement of \(\sum_{n=1}^\infty a_n\) satisfies \[{\color{purple}\liminf_{n \to \infty}s_n'=\alpha} \quad \text{ and }\quad {\color{brown}\limsup_{n \to \infty}s_n'=\beta}.\]
Choose \((\alpha_n)_{n \in \mathbb{N}}\) and \((\beta_n)_{n \in \mathbb{N}}\) so that \(\alpha_n<\beta_n\) with \(\beta_1>0\) and \[\alpha_n \ _{\overrightarrow{n \to \infty}}\ \alpha, \quad \text{ and } \quad \beta_n \ _{\overrightarrow{n \to \infty}}\ \beta.\]
Let \(m_1,k_1\) be the smallest integers such that \[{\color{red}P_1+\ldots+P_{m_1}}>\beta_1 \quad \text{ and }\quad {\color{blue}{\color{red}P_1+\ldots+P_{m_1}}-Q_1-\ldots-Q_{k_1}}<\alpha_1.\]
Let \(m_2,k_2\) be the smallest integers such that \[{\color{red}P_1+\ldots+P_{m_1}}{\color{blue}-Q_1-\ldots-Q_{k_1}}{\color{red}+P_{m_1+1}+\ldots+P_{m_2}}>\beta_2,\] and \[\begin{aligned} &{\color{red}P_1+\ldots+P_{m_1}}{\color{blue}-Q_1-\ldots-Q_{k_1}}\\ &\qquad {\color{red}+P_{m+1}+\ldots+P_{m_2}}{\color{blue}-Q_{k_1+1}-\ldots-Q_{k_2}}<\alpha_2. \end{aligned}\] and we continue this way.
This is possible since \[\left|\sum_{n=1}^{\infty}P_n \right|=\infty \quad \text{ and }\quad \left|\sum_{n=1}^{\infty}Q_n \right|=\infty.\]
If \(x_n,y_n\) are the partial sums of \[\begin{aligned} &{\color{red}P_1+\ldots+P_{m_1}}{\color{blue}-Q_1-\ldots-Q_{k_1}}\\ &\qquad {\color{red}+P_{m+1}+\ldots+P_{m_2}}{\color{blue}-Q_{k_1+1}-\ldots-Q_{k_2}} \end{aligned}\] whose last terms respectively are \(P_{m_n}\) and \(Q_{k_n}\) then \[|x_n-\beta_n| \leq P_{m_n}\quad \text{ and }\quad |y_n-\alpha_n| \leq Q_{k_n}.\]
Since \[\beta_n<x_n \leq (x_n-P_{m_n})+P_{m_n} \leq \beta_n+P_{m_n},\] then \[0<x_n-\beta_n\le P_{m_n}.\]
Since \(P_n \ _{\overrightarrow{n \to \infty}}\ 0\) and \(Q_{n} \ _{\overrightarrow{n \to \infty}}\ 0\) we see that \[{\color{blue}x_n \ _{\overrightarrow{n \to \infty}}\ \beta}.\]
Similarly we conclude that \[{\color{red}y_n \ _{\overrightarrow{n \to \infty}}\ \alpha}.\]
Finally it is clear that no number less that \(\alpha\) or greater that \(\beta\) can be subsequential limit of the partial sums of \(s_n'\).
This completes the proof of the theorem. $$\tag*{$\blacksquare$}$$
Theorem. If \(\sum_{n=1}^{\infty}|a_n|<\infty\), then every rearrangement of \(\sum_{n=1}^{\infty}a_n\) converge to the same limit.
Proof. Let \(\sum_{n=1}^{\infty}a_n'\) be a rearrangement of \(\sum_{n=1}^{\infty}a_n\) with partial sums \(s_n'\). Given \(\varepsilon>0\) there is \(N_{\varepsilon} \in \mathbb{N}\) such that \(m \geq n \geq N_{\varepsilon}\) implies \[\sum_{k=n}^m|a_k|<\varepsilon.\]
Now choose \(p \in \mathbb{N}\) such that \[\{1,2,\ldots,N_{\varepsilon}\} \subseteq \{k_1,k_2,\ldots,k_p\}; \quad \text{ here }\quad a_{n}'=a_{k_n}.\]
If \(n>p\) then the numbers \(a_1,\ldots,a_{N_{\varepsilon}}\) will cancel in the difference \(s_n-s_n'\) so that
\[|s_n-s_n'|<\varepsilon.\]
Hence \(s_n'\) converges to the same limit as \((s_n)_{n \in \mathbb{N}}\).$$\tag*{$\blacksquare$}$$