Amalie Karcher (1819-1887) was a German painter.
We have sections
- Infinite Series \ref{a}
- Inserting and Removing Parentheses
- Rearrangement of Series
- Tests for Convergence of Series
- Double Sequence and Series
\begin{equation}{\label{a}}\tag{A}\mbox{}\end{equation}
Infinite Series.
Let \(\{a_{n}\}_{n=1}^{\infty}\) be a sequence of real numbers. We can define a new sequence \(\{s_{n}\}_{n=1}^{\infty}\) given by
\[s_{n}=a_{1}+\cdots +a_{n}=\sum_{k=1}^{n}a_{k}.\]
The ordered pair \((\{a_{n}\}_{n=1}^{\infty},\{s_{n}\}_{n=1}^{\infty})\) is called an infinite series, and the number \(s_{n}\) is called the \(n\)th partial sum of the series. The series is said to be convergent when the sequence \(\{s_{n}\}_{n=1}^{\infty}\) of partial sums is convergent. Also, the infinite series is said to be divergent when the sequence \(\{s_{n}\}_{n=1}^{\infty}\) of partial sums is divergent.
\begin{equation}{\label{map161}}\tag{1}\mbox{}\end{equation}
Proposition \ref{map161}. Let \(a=\sum_{k=1}^{\infty}a_{n}\) and \(b=\sum_{k=1}^{\infty}b_{k}\) be two convergent series. Given any constants \(\alpha\) and \(\beta\), the series \(\sum_{k=1}^{\infty}(\alpha a_{k}+\beta b_{k})\) converges to \(\alpha a+\beta b\). In other words, we have
\[\sum_{k=1}^{\infty}\left (\alpha a_{k}+\beta b_{k}\right )=\alpha\sum_{k=1}^{\infty}a_{k}+\beta\sum_{k=1}^{\infty}b_{k}.\]
Proof. We define the sequence \(\{s_{n}\}_{n=1}^{\infty}\) as
\[s_{n}=\sum_{k=1}^{n}\left (\alpha a_{k}+\beta b_{k}\right )=\alpha\sum_{k=1}^{n}a_{k}+\beta\sum_{k=1}^{n}b_{k}.\]
By taking \(n\rightarrow\infty\), we obtain the desired results. \(\blacksquare\)
\begin{equation}{\label{map150}}\tag{2}\mbox{}\end{equation}
Proposition \ref{map150}. A monotonic sequence is convergent if and only if it is bounded.
\end{Pro}
\begin{Proof}
Suppose that \(\{a_{n}\}_{n=1}^{\infty}\) is increasing. Then, we have
\[\lim_{n\rightarrow\infty}a_{n}=\sup_{n}a_{n}.\]
Suppose that \(\{a_{n}\}_{n=1}^{\infty}\) is decreasing. Then, we have
\[\lim_{n\rightarrow\infty}a_{n}=\inf_{n}a_{n}.\]
This completes the proof. \(\blacksquare\)
\begin{equation}{\label{map162}}\tag{3}\mbox{}\end{equation}
Proposition \ref{map162}. Let \(\{a_{n}\}_{n=1}^{\infty}\) be a sequence with \(a_{n}\geq 0\) for all \(n\), and let \(\{s_{n}\}_{n=1}^{\infty}\) be the sequence of partial sums of the series \(\sum_{k=1}^{\infty}a_{k}\). Then, the series \(\sum_{k=1}^{\infty}a_{k}\) is convergent if and only if the sequence \(\{s_{n}\}_{n=1}^{\infty}\) is bounded.
Proof. Since \(s_{n}=a_{1}+\cdots +a_{n}\) and \(a_{n}\geq 0\) for all \(n\), it follows that \(\{s_{n}\}_{n=1}^{\infty}\) is an increasing sequence. The result follows immediately from Proposition~\ref{map150}. \(\blacksquare\)
\begin{equation}{\label{map266}}\tag{4}\mbox{}\end{equation}
Proposition \ref{map266}. Let \(\{a_{n}\}_{n=1}^{\infty}\) and \(\{b_{n}\}_{n=1}^{\infty}\) be two sequences satisfying \(a_{n}=b_{n+1}-b_{n}\) for all \(n\). Then, the series \(\sum_{k=1}^{\infty}a_{k}\) is convergent if and only if the limit \(\lim_{n\rightarrow\infty}b_{n}\) exists. Moreover, we have
\[\sum_{k=1}^{\infty}a_{k}=\lim_{n\rightarrow\infty}b_{n}-b_{1}.\]
Proof. The results follow by the following formula
\[\sum_{k=1}^{n}a_{k}=\sum_{k=1}^{n}(b_{k+1}-b_{k})=b_{n+1}-b_{1}\]
immediately. \(\blacksquare\)
\begin{equation}{\label{map151}}\tag{5}\mbox{}\end{equation}
Proposition \ref{map151}. (Cauchy condition for series of real numbers). The series \(\sum_{k=1}^{\infty}a_{k}\) is convergent if and only if, given any \(\epsilon >0\), there exists an integer \(N\) such that
\[n>N\mbox{ implies }\left |a_{n+1}+\cdots +a_{n+p}\right |<\epsilon\]
for all \(p\).
Proof. Let \(s_{n}=a_{1}+\cdots +a_{n}\). Therefore, we have
\[s_{n+p}-s_{n}=a_{n+1}+\cdots +a_{n+p}.\]
The result follows from the Cauchy condition for sequence of real numbers by referring to the page Limits of sequences. $\blacksquare$
\begin{equation}{\label{mac152}}\tag{6}\mbox{}\end{equation}
Corollary \ref{mac152}. Suppose that the series \(\sum_{k=1}^{\infty}a_{k}\) is convergent. Then, we have \(\lim_{n\rightarrow\infty}a_{n}=0\).
Proof. The result follows immediately from Proposition \ref{map151} by taking \(p=1\). \(\blacksquare\)
The converse of Corollary \ref{mac152} is not true. For example, the harmonic series \(\sum_{k=1}^{\infty}\frac{1}{k}\) diverges to \(+\infty\). However, we have \(1/n\rightarrow 0\).
\begin{equation}{\label{ma128}}\tag{7}\mbox{}\end{equation}
Proposition \ref{ma128}. Let \(\{a_{n}\}_{n=1}^{\infty}\) be a convergent sequence, and let \(\{\sigma_{n}\}_{n=1}^{\infty}\) be a sequence of arithmetic means given by
\[\sigma_{n}=\frac{1}{n}\sum_{k=1}^{n}a_{n}.\]
Then, we have that
\[\lim_{n\rightarrow\infty}a_{n}=A\mbox{ implies }\lim_{n\rightarrow\infty}\sigma_{n}=A.\]
Proof. Let \(b_{n}=a_{n}-A\) and \(\tau_{n}=\sigma_{n}-A\). Then, we have
\begin{align*} \tau_{n} & =\frac{1}{n}\sum_{k=1}^{n}a_{k}-A\\ & =\frac{1}{n}\left (\sum_{k=1}^{n}a_{k}-nA\right )
\\ & =\frac{1}{n}\sum_{k=1}^{n}\left (a_{k}-A\right )\\ & =\frac{1}{n}\sum_{k=1}^{n}b_{k}.\end{align*}
Since \(\lim_{n\rightarrow\infty}b_{n}=0\), given any \(\epsilon>0\), there exists an integer \(N\) such that \(n\geq N\) implies \(|b_{n}|<\epsilon\). Let
\[M=\max\left\{\left |b_{1}\right |,\left |b_{2}\right |,\cdots,\left |b_{N}\right |\right\}.\]
Then, for \(n>N\), we have
\[|\tau_{n}|\leq\frac{1}{n}\sum_{k=1}^{N}\left |b_{k}\right |+\frac{1}{n}\sum_{k=N+1}^{n}
\left |b_{k}\right |\leq\frac{NM}{n}+\frac{n-N}{n}\epsilon<\frac{NM}{n}+\epsilon,\]
which implies
\[0\leq\lim_{n\rightarrow\infty}|\tau_{n}|\leq\epsilon,\]
Since \(\epsilon\) can be any positive number, it follows \(\lim_{n\rightarrow\infty}\tau_{n}=0\). This completes the proof. \(\blacksquare\)
\begin{equation}{\label{ma129}}\tag{8}\mbox{}\end{equation}
Proposition \ref{ma129}. Suppose that the series \(\sum_{k=1}^{\infty}a_{k}\) is convergent to \(S\). Let \(s_{n}\) be the \(n\)th partial sums of \(\sum_{k=1}^{\infty}a_{k}\), and let \(\{\zeta_{n}\}_{n=1}^{\infty}\) be a sequence of arithmetic means of partial sums given by
\[\zeta_{n}=\frac{1}{n}\sum_{k=1}^{n}s_{n}.\]
Then, we have
\[\lim_{n\rightarrow\infty}\zeta_{n}=S=\sum_{k=1}^{\infty}a_{k}.\]
Proof. Since \(\lim_{n\rightarrow\infty}s_{n}=S\), the desired result follows from Proposition \ref{ma128} immediately. \(\blacksquare\)
\begin{equation}{\label{b}}\tag{B}\mbox{}\end{equation}
Inserting and Removing Parentheses.
Let \(p\) be a function defined on \(\mathbb{N}\) whose range is a subset of \(\mathbb{N}\) such that \(n<m\) implies \(p(n)<p(m)\) with \(n\leq p(n)\) and \(m\leq p(m)\). Given a series \(\sum_{k=1}^{\infty}a_{k}\), we define a new series \(\sum_{k=1}^{\infty}b_{k}\) given by
\[b_{1}=a_{1}+a_{2}+\cdots +a_{p(1)}\]
and
\[b_{n+1}=a_{p(n)+1}+a_{p(n)+2}+\cdots +a_{p(n+1)}\mbox{ for all }n.\]
Then, we say that \(\sum_{k=1}^{\infty}b_{k}\) is obtained from \(\sum_{k=1}^{\infty}a_{k}\) by inserting parentheses, and that \(\sum_{k=1}^{\infty}a_{k}\) is obtained from \(\sum_{k=1}^{\infty}b_{k}\) by removing parentheses. For example, if \(p(1)=3\) and \(p(2)=5\), we have
\[b_{1}+b_{2}+\cdots=(a_{1}+a_{2}+a_{3})+(a_{4}+a_{5})+\cdots .\]
\begin{equation}{\label{map155}}\tag{9}\mbox{}\end{equation}
Proposition \ref{map155}. Suppose that the series \(\sum_{k=1}^{\infty}a_{k}\) converges to \(s\). Then, every series \(\sum_{k=1}^{\infty}b_{k}\) obtained from \(\sum_{k=1}^{\infty}a_{k}\) by inserting parentheses also converges to \(s\).
Proof. We write
\[s_{n}=\sum_{k=1}^{n}a_{k}\mbox{ and }t_{n}=\sum_{k=1}^{n}b_{k}.\]
Then \(\{t_{n}\}_{n=1}^{\infty}\) is a subsequence of \(\{s_{n}\}_{n=1}^{\infty}\) satisfying \(t_{n}=s_{p(n)}\). Since the sequence \(\{s_{n}\}_{n=1}^{\infty}\) is convergent, the subsequence \(\{t_{n}\}_{n=1}^{\infty}\) is also convergent, and the proof is complete. \(\blacksquare\)
Removing parentheses may destroy convergence. For example, the series
\[0+0+\cdots =(1-1)+(1-1)+\cdots\mbox{ converges to }0.\]
However, after removing the parentheses, the series
\[1-1+1-1+\cdots\]
is obvious not convergent.
\begin{equation}{\label{map156}}\tag{10}\mbox{}\end{equation}
Proposition \ref{map156}. Let the series \(\sum_{k=1}^{\infty}a_{k}\) be obtained from \(\sum_{k=1}^{\infty}b_{k}\) by removing parentheses. In other words, the series \(\sum_{k=1}^{\infty}b_{k}\) is obtained from \(\sum_{k=1}^{\infty}a_{k}\) by inserting parentheses. Suppose that there exists a constant \(M>0\) satisfying \(p(n+1)-p(n)<M\) for all \(n\), and that \(\lim_{n\rightarrow\infty}a_{n}=0\). Then, the series \(\sum_{k=1}^{\infty}a_{k}\) is convergent if and only if the series \(\sum_{k=1}^{\infty}b_{k}\) is convergent. Moreover, these two series converge to the same value.
Proof. Proposition \ref{map155} says that we only need to prove the converse. Suppose that the series \(\sum_{k=1}^{\infty}b_{k}\) is convergent. We define
\[\begin{array}{llcl}
{\displaystyle s_{n}=\sum_{k=1}^{n}a_{k}}, & {\displaystyle t_{n}=\sum_{k=1}^{n}b_{k}},
& \mbox{and} & {\displaystyle t=\lim_{n\rightarrow\infty}t_{n}}.
\end{array}\]
Since
\[\lim_{n\rightarrow\infty}a_{n}=0\mbox{ and }t=\lim_{n\rightarrow\infty}t_{n},\]
given \(\epsilon >0\), there exists an integer \(N\) such that
\begin{equation}{\label{maeq156}}\tag{11}
n>N\mbox{ implies }\left |t_{n}-t\right |<\frac{\epsilon}{2}\mbox{ and }\left |a_{n}\right |<\frac{\epsilon}{2M}.
\end{equation}
For \(m\geq N\), we consider
\begin{align*}
& \cdots +b_{N}+\cdots +b_{m}+b_{m+1}+\cdots\\
& \quad =\cdots +\left (a_{p(N-1)+1}+\cdots +a_{p(N)}\right )+\cdots\\
& \quad\quad+\left (a_{p(m-1)+1}+\cdots +a_{p(m)}\right )+\left (a_{p(m)+1}+\cdots +a_{p(m+1)}\right )+\cdots.
\end{align*}
Given \(a_{n}>a_{p(N)}\), we can find \(m\geq N\) satisfying \(a_{p(m)}\leq a_{n}<a_{p(m+1)}\). For such \(n\) and \(m\), we have
\begin{align*}
s_{n} & =a_{1}+\cdots +a_{p(N)}+\cdots +a_{p(m)}+a_{p(m)+1}+\cdots +a_{n}\\
& =a_{1}+\cdots +a_{p(N)}+\cdots +a_{p(m)}+a_{p(m)+1}+\cdots+a_{n}+a_{n+1}+\cdots +a_{p(m+1)}\\
& \quad -\left (a_{n+1}+a_{n+2}+\cdots +a_{p(m+1)}\right )\\
& =b_{1}+\cdots +b_{m+1}-\left (a_{n+1}+a_{n+2}+\cdots +a_{p(m+1)}\right )\\
& =t_{m+1}-\left (a_{n+1}+a_{n+2}+\cdots +a_{p(m+1)}\right ).
\end{align*}
Therefore, we obtain
\begin{align*}
\left |s_{n}-t\right | & \leq\left |t_{m+1}-t\right |+\left |a_{n+1}+a_{n+2}+\cdots +a_{p(m+1)}\right |\\
& \leq\left |t_{m+1}-t\right |+\left |a_{p(m)+1}+a_{p(m)+2}+\cdots +a_{n}\right |+\left |a_{n+1}+a_{n+2}+\cdots +a_{p(m+1)}\right |\\
& \leq\left |t_{m+1}-t\right |+\left |a_{p(m)+1}\right |+\left |a_{p(m)+2}\right |+\cdots +\left |a_{p(m+1)}\right |\\
& <\frac{\epsilon}{2}+\frac{\epsilon}{2M}\left [p(m+1)-p(m)\right ]\mbox{ (by (\ref{maeq156}))}\\
& <\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon .
\end{align*}
This proves
\[\lim_{n\rightarrow\infty}s_{n}=t,\]
and the proof is complete. \(\blacksquare\)
Definition. Suppose that \(a_{n}>0\) for all \(n\). Then, the the series
\[\sum_{k=1}^{\infty}(-1)^{k+1}a_{k}\]
is called an alternating series. \(\sharp\)
\begin{equation}{\label{map157}}\tag{12}\mbox{}\end{equation}
Proposition \ref{map157}. Let \(\{a_{n}\}_{n=1}^{\infty}\) be a decreasing sequence with \(a_{n}>0\) for all \(n\). Suppose that \(\lim_{n\rightarrow\infty}a_{n}=0\). Then, the alternating series \(\sum_{k=1}^{\infty}(-1)^{k+1}a_{k}\) is convergent.
Proof. We are going to insert the parentheses to the alternating series by grouping together two terms at a time. Therefore, we take \(p(n)=2n\). Then, the new series \(\sum_{k=1}^{\infty}b_{k}\) is given by
\[\begin{array}{llcl}
b_{1}=a_{1}-a_{2}, & b_{2}=a_{3}-a_{4}, & \cdots ,& b_{n}=a_{2n-1}-a_{2n}
\end{array}\]
Since
\[\lim_{n\rightarrow\infty}a_{n}=0\mbox{ and }p(n+1)-p(n)=2,\]
Proposition \ref{map156} says that if the series \(\sum_{k=1}^{\infty}b_{k}\) is convergent, then the alternating series is also convergent. Since \(\{a_{n}\}_{n=1}^{\infty}\) is decreasing, we have \(b_{n}\geq 0\) for all \(n\). Since
\[a_{2n}\geq 0,\quad a_{2}-a_{3}\geq 0,\cdots ,\quad a_{2n-2}-a_{2n-1}\geq 0\]
by the decreasing assumption of \(\{a_{n}\}_{n=1}^{\infty}\), the following partial sums are bounded above by \(a_{1}\)
\[\sum_{k=1}^{n}b_{k}=a_{1}-(a_{2}-a_{3})-\cdots -(a_{2n-2}-a_{2n-1})-a_{2n}<a_{1}.\]
This shows that the series \(\sum_{k=1}^{\infty}b_{k}\) is convergent by Proposition \ref{map162}, i.e., the alternating series is also convergent. This completes the proof. \(\blacksquare\)
Definition. A series \(\sum_{k=1}^{\infty}a_{k}\) is called absolutely convergent when the series \(\sum_{k=1}^{\infty}|a_{k}|\) is convergent. Suppose that the series \(\sum_{k=1}^{\infty}a_{k}\) is convergent such that \(\sum_{k=1}^{\infty}|a_{k}|\) is divergent. Then, it is called conditionally convergent. \(\sharp\)
Proposition. Suppose that the series \(\sum_{k=1}^{\infty}a_{k}\) is absolutely convergent. Then, it is also convergent.
Proof. Applying the Cauchy condition presented in Proposition \ref{map151} to the following inequality
\[\left |a_{n+1}+\cdots +a_{n+p}\right |\leq\left ||a_{n+1}|+\cdots +|a_{n+p}|\right |,\]
the proof is complete. \(\blacksquare\)
Example. Using Proposition \ref{map157}, the alternating series
\[\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{n}\]
is convergent. However, the series
\[\sum_{k=1}^{\infty}\left |\frac{(-1)^{k+1}}{n}\right |=\sum_{k=1}^{\infty}\frac{1}{n}\]
diverges to \(+\infty\). \(\sharp\)
Proposition. Given a series \(\sum_{k=1}^{\infty}a_{k}\), we define
\begin{equation}{\label{maeq159}}\tag{13}
p_{n}=\frac{|a_{n}|+a_{n}}{2}\mbox{ and }q_{n}=\frac{|a_{n}|-a_{n}}{2}.
\end{equation}
Then, we have the following properties.
(i) Suppose that the series \(\sum_{k=1}^{\infty}a_{k}\) is conditionally convergent. Then, both series \(\sum_{k=1}^{\infty}p_{k}\) and \(\sum_{k=1}^{\infty}q_{k}\) are divergent.
(ii) Suppose that the series \(\sum_{k=1}^{\infty}a_{k}\) is absolutely convergent. Then, both series \(\sum_{k=1}^{\infty}p_{k}\) and \(\sum_{k=1}^{\infty}q_{k}\) are convergent, and we have
\begin{equation}{\label{maeq160}}\tag{14}
\sum_{k=1}^{\infty}a_{k}=\sum_{k=1}^{\infty}p_{k}-\sum_{k=1}^{\infty}q_{k}.
\end{equation}
Proof. We have
\[a_{n}=p_{n}-q_{n}\mbox{ and }|a_{n}|=p_{n}+q_{n}.\]
To prove part (i), we assume that the series \(\sum_{k=1}^{\infty}a_{k}\) is convergent, and that the series \(\sum_{k=1}^{\infty}|a_{k}|\) is divergent. Since \(p_{n}=a_{n}+q_{n}\), if \(\sum_{k=1}^{\infty}q_{k}\) is convergent, then \(\sum_{k=1}^{\infty}p_{k}\) is also convergent by Proposition \ref{map161}. Similarly, if \(\sum_{k=1}^{\infty}p_{k}\) is convergent, then \(\sum_{k=1}^{\infty}q_{k}\) is also convergent. Therefore, if either \(\sum_{k=1}^{\infty}q_{k}\) or \(\sum_{k=1}^{\infty}p_{k}\) is convergent, then both series \(\sum_{k=1}^{\infty}q_{k}\) and \(\sum_{k=1}^{\infty}p_{k}\) are convergent, which implies that \(\sum_{k=1}^{\infty}|a_{k}|\) is convergent, since \(|a_{n}|=p_{n}+q_{n}\). This contradiction says that both series \(\sum_{k=1}^{\infty}p_{k}\) and \(\sum_{k=1}^{\infty}q_{k}\) are divergent.
To prove part (ii), since both series \(\sum_{k=1}^{\infty}a_{k}\) and \(\sum_{k=1}^{\infty}|a_{k}|\) are convergent, from (\ref{maeq159}), we see that both series \(\sum_{k=1}^{\infty}q_{k}\) and \(\sum_{k=1}^{\infty}p_{k}\) are convergent. Finally, the equality (\ref{maeq160}) follows from \(p_{n}=a_{n}+q_{n}\) and Proposition \ref{map161}. This completes the proof. \(\blacksquare\)
\begin{equation}{\label{c}}\tag{C}\mbox{}\end{equation}
Rearrangement of Series.
Suppose that the function \(f:\mathbb{N}\rightarrow\mathbb{N}\) is one-to-one. Let \(\sum_{k=1}^{\infty}a_{k}\) and \(\sum_{k=1}^{\infty}b_{k}\) be two series satisfying \(b_{n}=a_{f(n)}\) for all \(n\). Then \(\sum_{k=1}^{\infty}b_{k}\) is said to be a rearrangement of \(\sum_{k=1}^{\infty}a_{k}\). Since \(f\) is one-to-one, we also have \(a_{n}=b_{f^{-1}(n)}\) for all \(n\). This says that \(\sum_{k=1}^{\infty}a_{k}\) is also a rearrangement of \(\sum_{k=1}^{\infty}b_{k}\).
Proposition. Suppose that \(\sum_{k=1}^{\infty}a_{k}\) is absolutely convergent to \(s\). Then, every rearrangement of \(\sum_{k=1}^{\infty}a_{k}\) is also absolutely convergent to \(s\).
Proof. Let \(b_{n}=a_{f(n)}\) for all \(n\). Then, we have
\[\sum_{k=1}^{n}|b_{k}|\leq\sum_{k=1}^{n}|a_{f(k)}|\leq\sum_{k=1}^{\infty}|a_{k}|<+\infty,\]
which says that the partial sum \(\sum_{k=1}^{n}|b_{k}|\) is bounded above by \(\sum_{k=1}^{\infty}a_{k}\). It means that the series \(\sum_{k=1}^{\infty}b_{k}\) is absolutely convergent by Proposition \ref{map161}. Now, we are going to show that \(\sum_{k=1}^{\infty}b_{k}=s\). Let
\[t_{n}=\sum_{k=1}^{n}b_{k}\mbox{ and }s_{n}=\sum_{k=1}^{n}a_{k}.\]
Since \(\sum_{k=1}^{\infty}a_{k}\) is absolutely convergent to \(s\), using Cauchy condition, given any \(\epsilon>0\), there exists an integer \(N\) satisfying
\begin{equation}{\label{ma115}}\tag{15}
|s_{N}-s|<\frac{\epsilon}{2}\mbox{ and }\sum_{k=1}^{p}|a_{N+k}|<\frac{\epsilon}{2}
\end{equation}
for sufficiently large \(p\). Therefore, we obtain
\begin{equation}{\label{ma114}}\tag{16}
|t_{n}-s|\leq |t_{n}-s_{N}|+|s_{N}-s|<|t_{n}-s_{N}|+\frac{\epsilon}{2}.
\end{equation}
We can take a sufficiently large integer \(M\) satisfying
\[\{1,2,\cdots,N\}\subseteq\{f(1),f(2),\cdots,f(M)\}.\]
Then, using (\ref{ma115}), we see that \(n>M\) implies \(f(n)>N\) and
\begin{align*}
|t_{n}-s_{N}| & =\left |b_{1}+\cdots+b_{n}-\left (a_{1}+\cdots+a_{N}\right )\right |\\
& =\left |a_{f(1)}+\cdots+a_{f(n)}-\left (a_{1}+\cdots+a_{N}\right )\right |\\
& \leq\left |a_{N+1}\right |+\left |a_{N+2}\right |+\cdots +a_{N+p}\\
& =\sum_{k=1}^{p}|a_{N+k}|<\frac{\epsilon}{2}\mbox{ for sufficiently large }p.
\end{align*}
Using (\ref{ma114}), we conclude that
\[n>M\mbox{ implies }|t_{n}-s|<\epsilon.\]
This completes the proof. \(\blacksquare\)
\begin{equation}{\label{ma116}}\tag{17}\mbox{}\end{equation}
Theorem \ref{ma116}. Suppose that \(\sum_{k=1}^{\infty}a_{k}\) is a conditionally convergent series. Given any \(x\) and \(y\) in \([-\infty ,+\infty ]\) satisfying \(x\leq y\), there exists a rearrangement \(\sum_{k=1}^{\infty}b_{k}\) of \(\sum_{k=1}^{\infty}a_{k}\) satisfying
\[\liminf_{n\rightarrow\infty}t_{n}=x\mbox{ and }\limsup_{n\rightarrow\infty}t_{n}=y,\]
where \(t_{n}=b_{1}+\cdots +b_{n}\) is the partial sum of \(\sum_{k=1}^{\infty}b_{k}\). \(\sharp\)
Suppose that we consider \(x=y\) in Theorem \ref{ma116}. Then, we have
\[\liminf_{n\rightarrow\infty}t_{n}=x=y\limsup_{n\rightarrow\infty}t_{n},\]
which says
\[\lim_{n\rightarrow\infty}t_{n}=x=y.\]
In other words, given any \(x\in\mathbb{R}\), there exists a rearrangement \(\sum_{k=1}^{\infty}b_{k}\) of \(\sum_{k=1}^{\infty}a_{k}\) satisfying
\[\sum_{k=1}^{\infty}b_{k}=x.\]
Let \(\mathbb{S}\) be an infinite subset of \(\mathbb{N}\). Suppose that the function \(f:\mathbb{N}\rightarrow\mathbb{S}\) is one-to-one. Let \(\sum_{k=1}^{\infty}a_{k}\) and \(\sum_{k=1}^{\infty}b_{k}\) be two series satisfying \(b_{n}=a_{f(n)}\) for all \(n\). Then \(\sum_{k=1}^{\infty}b_{k}\) is said to be a subseries of \(\sum_{k=1}^{\infty}a_{k}\). For example, we take \(f(n)=2n\). Then
\[\sum_{k=1}^{\infty}b_{k}=\sum_{k=1}^{\infty}a_{2k}=a_{2}+a_{4}+\cdots\]
is a subseries of \(\sum_{k=1}^{\infty}a_{k}\).
\begin{equation}{\label{ma117}}\tag{18}\mbox{}\end{equation}
Proposition \ref{ma117}. Suppose that the series \(\sum_{k=1}^{\infty}a_{k}\) is absolutely convergent. Then, every subseries \(\sum_{k=1}^{\infty}b_{k}\) of \(\sum_{k=1}^{\infty}a_{k}\) is also absolutely convergent. Moreover, we have
\[\left |\sum_{k=1}^{\infty}b_{k}\right |\leq\sum_{k=1}^{\infty}|b_{k}|\leq\sum_{k=1}^{\infty}|a_{k}|.\]
Proof. Given any integer \(n\), let
\[N=\max\left\{f(1),\cdots,f(n)\right\}.\]
Then, we have
\[\left |\sum_{k=1}^{n}b_{k}\right |\leq\sum_{k=1}\left |b_{k}\right |\leq\sum_{k=1}^{N}
\left |a_{k}\right |\leq\sum_{k=1}^{\infty}\left |a_{k}\right |.\]
The absolute convergence of \(\sum_{k=1}^{\infty}a_{k}\) says that the partial sums \(\sum_{k=1}\left |b_{k}\right |\) is bounded above. This shows that \(\sum_{k=1}^{\infty}b_{k}\) is also absolutely convergent. \(\blacksquare\)
Proposition. Let \(\{f_{1},f_{2},\cdots\}\) be a countable collection of functions such that each of them is defined on \(\mathbb{N}\). Suppose that the following conditions are satisfied.
- Each \(f_{n}\) is one-to-one.
- The range \(Q_{n}\equiv f_{n}(\mathbb{N})\) is a subset of \(\mathbb{N}\).
- \(\{Q_{1},Q_{2},\cdots\}\) is a collection of disjoint sets satisfying \(\bigcup_{k=1}^{\infty}Q_{k}=\mathbb{N}\).
Let \(\sum_{k=1}^{\infty}a_{k}\) be an absolutely convergent series. Then, we have the following properties.
(i) For each \(k\), the subseries \(\sum_{n=1}^{\infty}a_{f_{k}(n)}\) of \(\sum_{k=1}^{\infty}a_{k}\) is absolutely convergent.
(ii) We define \(s_{k}=\sum_{n=1}^{\infty}a_{f_{k}(n)}\). Then, the series \(\sum_{k=1}^{\infty}s_{k}\) is absolutely convergent and
\[\sum_{k=1}^{\infty}s_{k}=\sum_{k=1}^{\infty}a_{k}.\]
Proof. Part (i) follows immediately from Proposition \ref{ma117}. To prove part (ii), let
\[t_{k}=\left |s_{1}\right |+\cdots+\left |s_{k}\right |.\]
Then, we have
\begin{align*}
t_{k} & \leq\sum_{n=1}^{\infty}\left |b_{1}(n)\right |+\cdots+\sum_{n=1}^{\infty}\left |b_{k}(n)\right |\\ & =\sum_{n=1}^{\infty}\left [\left |b_{1}(n)\right |+\cdots+\left |b_{k}(n)\right |\right ]\\
& =\sum_{n=1}^{\infty}\left [\left |a_{f_{1}(n)}\right |+\cdots+\left |a_{f_{k}(n)}\right |\right ]\\ & \leq\sum_{n=1}^{\infty}\left |a_{n}\right |,
\end{align*}
The absolute convergence of \(\sum_{k=1}^{\infty}a_{k}\) says that the the partial sums \(\{t_{k}\}_{k=1}^{\infty}\) is bounded above. This shows that the series \(\sum_{k=1}^{\infty}s_{k}\) is absolutely convergent. Now, we want to find the sum of \(\sum_{k=1}^{\infty}s_{k}\). Given any \(\epsilon>0\), there exists an integer \(N\) such that
\[n\geq N\mbox{ implies }0<\sum_{k=1}^{\infty}\left |a_{k}\right |-\sum_{k=1}^{n}\left |a_{k}\right |<\frac{\epsilon}{2}.\]
We also have that \(n\geq N\) implies
\[\left |a_{N+1}\right |+\left |a_{N+2}\right |+\cdots <\frac{\epsilon}{2}\]
and
\begin{equation}{\label{ma118}}\tag{19}
\left |\sum_{k=1}^{\infty}a_{k}-\sum_{k=1}^{n}a_{k}\right |\leq\left |\sum_{k=1}^{\infty}\left |a_{k}\right |-\sum_{k=1}^{n}\left |a_{k}\right |\right |
<\frac{\epsilon}{2}.
\end{equation}
We can take many enough functions \(f_{1},\cdots, f_{r}\) such that each term \(a_{1},\cdots,a_{N}\) will appear somewhere in the following sum
\[\sum_{k=1}^{\infty}a_{f_{1}(k)}+\cdots+\sum_{k=1}^{\infty}a_{f_{r}(k)},\]
where the integer \(r\) depends on \(N\) and \(\epsilon\). For \(n>r\) and \(n>N\), we have
\begin{align}
\left |\sum_{k=1}^{n}s_{k}-\sum_{k=1}^{n}a_{k}\right |
& =\left |\sum_{k=1}^{\infty}a_{f_{1}(k)}+\cdots+\sum_{k=1}^{\infty}a_{f_{n}(k)}-\sum_{k=1}^{n}a_{k}\right |\nonumber\\
& \leq\left |a_{N+1}\right |+\left |a_{N+2}\right |+\cdots <\frac{\epsilon}{2},\label{ma119}\tag{20}
\end{align}
since the terms \(a_{1},\cdots,a_{N}\) are canceled out in the substraction. Therefore, using (\ref{ma118}) and (\ref{ma119}), we obtain
\begin{align*} \left |\sum_{k=1}^{n}s_{k}-\sum_{k=1}^{\infty}a_{k}\right | & \leq\left |\sum_{k=1}^{n}s_{k}-\sum_{k=1}^{n}a_{k}\right |+
\left |\sum_{k=1}^{n}a_{k}-\sum_{k=1}^{\infty}a_{k}\right |\\ & <\frac{\epsilon}{2}+\frac{\epsilon}{2}.\end{align*}
This completes the proof. \(\blacksquare\)
\begin{equation}{\label{d}}\tag{D}\mbox{}\end{equation}
Tests for Convergence of Series.
Test for convergence of infinite series is an important issue. For example, when we make sure that the infinite series is convergent, we can calculate its approximated sum for sufficiently many terms by using computers. We are going to provide many methods to test the convergence of given infinite series.
\begin{equation}{\label{mat163}}\tag{21}\mbox{}\end{equation}
Theorem \ref{mat163}. (Comparison Test). Given \(a_{n}>0\) and \(b_{n}>0\) for all \(n\), suppose that there exist positive constants \(c\) and \(N\) satisfying \(a_{n}\leq cb_{n}\) for all \(n\geq N\). Then, the convergence of series \(\sum_{k=1}^{\infty}b_{k}\) implies the convergence of series \(\sum_{k=1}^{\infty}a_{k}\).
Proof. Proposition \ref{map162} says that the partial sums of \(\sum_{k=1}^{\infty}b_{k}\) are bounded. Therefore, the partial sums of \(\sum_{k=1}^{\infty}a_{k}\) are also bounded. This shows that the series \(\sum_{k=1}^{\infty}a_{k}\) is convergent, and the proof is complete. \(\blacksquare\)
\begin{equation}{\label{mat164}}\tag{22}\mbox{}\end{equation}
Theorem \ref{mat164}. (Limit Comparison Test). Given \(a_{n}>0\) and \(b_{n}>0\) for all \(n\). Suppose that
\[\lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}=c>0.\]
Then, the series \(\sum_{k=1}^{\infty}a_{k}\) is convergent if and only if the series \(\sum_{k=1}^{\infty}b_{k}\) is convergent. Suppose that
\[\lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}=0.\]
Then, the convergence of series \(\sum_{k=1}^{\infty}b_{k}\) implies the convergence of series \(\sum_{k=1}^{\infty}a_{k}\).
Proof. For \(c>0\), given any \(\epsilon >0\) satisfying \(c-\epsilon >0\), the limit says that there exists an integer \(N\) such that
\[n\geq N\mbox{ implies }0<c-\epsilon <\frac{a_{n}}{b_{n}}<c+\epsilon .\]
Using Theorem \ref{mat163}, we obtain the desired result. For \(c=0\), given any \(\epsilon >0\), there exists an integer \(N\) such that
\[n\geq N\mbox{ implies }0<\frac{a_{n}}{b_{n}}<\epsilon .\]
Using Theorem~\ref{mat163} again, we conclude that the convergence of series \(\sum_{k=1}^{\infty}b_{k}\) implies the convergence of series \(\sum_{k=1}^{\infty}a_{k}\). This completes the proof. \(\blacksquare\)
\begin{equation}{\label{ma152}}\tag{23}\mbox{}\end{equation}
Example \ref{ma152}. Given a sequence \(\{I_{n}\}_{n=1}^{\infty}\) by
\[I_{n}=\int_{0}^{\pi/2}\cos^{n}xdx,\]
which is known as Wallis integral, it is clear to see
\[I_{0}=\frac{\pi}{2}\mbox{ and }I_{1}=1.\]
For \(n\geq 2\), using the integration by parts, we have
\begin{align*}
I_{n+2} & =\int_{0}^{\pi/2}\cos^{n+1}x\cos xdx\\ & =\left [\cos^{n+1}x\sin x\right ]_{0}^{\pi/2}+(n+1)\int_{0}^{\pi/2}\cos^{n}x\sin^{2}xdx\\
& =(n+1)\left (\int_{0}^{\pi/2}\cos^{n}xdx-\int_{0}^{\pi/2}\cos^{n+2}xdx\right )\\ & =(n+1)\left (I_{n}-I_{n+2}\right ),
\end{align*}
which implies
\begin{equation}{\label{ma151}}\tag{24}
I_{n+2}=\frac{n+1}{n+2}I_{n}.
\end{equation}
Using induction, we can obtain
\begin{align*} I_{2n} & =\frac{2n-1}{2n}\cdot\frac{2n-3}{2n-2}\cdots\frac{1}{2}\cdot I_{0}
\\ & =\frac{2n-1}{2n}\cdot\frac{2n-3}{2n-2}\cdots\frac{1}{2}\cdot\frac{\pi}{2}
\\ & =\frac{(2n)!\pi}{2^{2n+1}(n!)^{2}}\end{align*}
and
\begin{align*} I_{2n+1} & =\frac{2n}{2n+1}\cdot\frac{2n-2}{2n-1}\cdots\frac{2}{3}\cdot I_{1}
\\ & =\frac{2n}{2n+1}\cdot\frac{2n-2}{2n-1}\cdots\frac{2}{3}
\\ & =\frac{2^{2n}(n!)^{2}}{(2n+1)!}.\end{align*}
Since
\[0\leq\cos^{n+1}x\leq\cos^{n}x\mbox{ for }0\leq x\leq\pi/2,\]
it follows
\[I_{n+2}\leq I_{n+1}\leq I_{n}.\]
Since \(I_{n}>0\), we obtain
\[1\leq\frac{I_{n+1}}{I_{n+2}}\leq\frac{I_{n}}{I_{n+2}}=\frac{n+2}{n+1},\]
which implies
\[\lim_{n\rightarrow\infty}\frac{I_{n+1}}{I_{n}}=1.\]
From (\ref{ma151}), we have
\[(n+2)I_{n+2}I_{n+1}=(n+1)I_{n+1}I_{n},\]
which implies
\begin{align*} (n+2)I_{n+2}I_{n+1} & =(n+1)I_{n+1}I_{n}=nI_{n}I_{n-1}\\ & =\cdots=I_{0}I_{1}=\frac{\pi}{2}.\end{align*}
Therefore, we obtain
\begin{align*} \lim_{n\rightarrow\infty}nI_{n}^{2} & =\lim_{n\rightarrow\infty}(n+1)I_{n+1}^{2}
\\ & =\lim_{n\rightarrow\infty}(n+1)I_{n+1}I_{n}=\frac{\pi}{2},\end{align*}
which implies
\[\lim_{n\rightarrow\infty}2nI_{n}^{2}=\pi\mbox{ or }\lim_{n\rightarrow\infty}\sqrt{2n}I_{n}=\sqrt{\pi}.\]
Example. Given a sequence \(\{x_{n}\}_{n=1}^{\infty}\) by
\[x_{n}=\frac{n!}{\sqrt{n}}\left (\frac{e}{n}\right )^{n},\]
we are going to claim that the sequence \(\{x_{n}\}_{n=1}^{\infty}\) is convergent. Since \(x_{n}>0\) for all \(n\), we have
\begin{align*} \ln\left (\frac{x_{n+1}}{x_{n}}\right ) & =\ln\left (\frac{(n+1)!}{n!}\cdot\sqrt{\frac{n}{n+1}}
\cdot e\cdot\frac{n^{n}}{(n+1)^{n+1}}\right )\\ & =1-\left (n+\frac{1}{2}\right )\cdot\ln\left (1+\frac{1}{n}\right ).\end{align*}
Using the Taylor series of \(\ln (1+x)\), we have
\[\ln\left (1+\frac{1}{n}\right )=\frac{1}{n}-\frac{1}{2n^{2}}+\frac{1}{6n^{3}}+\frac{\epsilon_{n}}{n^{3}},\]
where \(\epsilon_{n}\rightarrow 0\) as \(n\rightarrow\infty\). Therefore, we obtain
\begin{align*} \ln\left (\frac{x_{n+1}}{x_{n}}\right ) & =1-\left (n+\frac{1}{2}\right )\cdot\ln\left (\frac{1}{n}-\frac{1}{2n^{2}}+\frac{1}{6n^{3}}+
\frac{\epsilon_{n}}{n^{3}}\right )\\ & =-\frac{1}{6n^{2}}+\frac{1}{4n^{2}}-\frac{\epsilon_{n}}{n^{2}}-\frac{\epsilon_{n}}{2n^{3}},\end{align*}
which implies
\[\lim_{n\rightarrow\infty}n^{2}\ln\left (\frac{x_{n+1}}{x_{n}}\right )=-\frac{1}{6}+\frac{1}{4}.\]
Using the limit comparison test, since the series \(\sum_{n=1}^{\infty}(1/n^{2})\) is convergent, it follows that the series
\[\sum_{n=1}^{\infty}\ln\left (\frac{x_{n+1}}{x_{n}}\right )=\sum_{n=1}^{\infty}\left [\ln(x_{n+1})-\ln (x_{n})\right ]\]
is convergent. This also says that the sequence \(\{s_{n}\}_{n=1}^{\infty}\) of its partial sums is convergent, where
\[s_{n}=\sum_{k=1}^{n}\left [\ln(x_{k+1})-\ln (x_{k})\right ]=\ln (x_{n+1})-\ln(x_{1}).\]
This shows that the sequence \(\{\ln(x_{n})\}_{n=1}^{\infty}\) is convergent, which also implies the sequence \(\{x_{n}\}_{n=1}^{\infty}\) is convergent.
Next, we are going to calculate its limit. Let
\[L=\lim_{n\rightarrow\infty}x_{n}=\lim_{n\rightarrow\infty}\frac{n!}{\sqrt{n}}\left (\frac{e}{n}\right )^{n}.\]
For sufficient large \(n\), we have
\[L\approx\frac{n!}{\sqrt{n}}\left (\frac{e}{n}\right )^{n},\]
which implies
\begin{equation}{\label{ma153}}\tag{25}
n!\approx L\cdot\left (\frac{n}{e}\right )^{n}\sqrt{n}.
\end{equation}
From Example \ref{ma152}, we have
\[I_{2n}=\frac{(2n)!\pi}{2^{2n+1}(n!)^{2}}\mbox{ and }\lim_{n\rightarrow\infty}\sqrt{2n}I_{n}=\sqrt{\pi}.\]
For sufficiently large \(n\), we also have
\[I_{n}\approx\sqrt{\frac{\pi}{2n}},\mbox{ i.e., }I_{2n}\approx\sqrt{\frac{\pi}{4n}},\]
which implies
\[\frac{(2n)!\pi}{2^{2n+1}(n!)^{2}}\approx\sqrt{\frac{\pi}{4n}}\mbox{ for sufficiently large }n.\]
Using (\ref{ma153}), we obtain
\[\frac{L\cdot\left (\frac{2n}{e}\right )^{2n}\sqrt{2n}\cdot\pi}{2^{2n+1}(L\cdot\left (\frac{n}{e}\right )^{n}\sqrt{n})^{2}}\approx\sqrt{\frac{\pi}{4n}}
\mbox{ for sufficiently large }n;\]
that is,
\[\frac{\pi}{2L}\cdot\frac{\left (\frac{2n}{e}\right )^{2n}\sqrt{2n}}{2^{2n}\cdot\left (\frac{n}{e}\right )^{2n}\sqrt{n^{2}}}\approx\sqrt{\frac{\pi}{4n}}
\mbox{ for sufficiently large }n,\]
which implies
\[\frac{\pi}{2L}\cdot\sqrt{\frac{2}{n}}\approx\sqrt{\frac{\pi}{4n}}\mbox{ for sufficiently large }n.\]
Therefore, we can solve
\[\lim_{n\rightarrow\infty}x_{n}=L=\sqrt{2\pi}.\]
Using (\ref{ma153}), we obtain the Stirling formula
\[n!\approx\sqrt{2\pi}\cdot\left (\frac{n}{e}\right )^{n}\sqrt{n}
=\sqrt{2n\pi}\cdot\left (\frac{n}{e}\right )^{n}\mbox{ for sufficiently large }n.\]
In order to use the comparison tests, we must have some series whose behavior are known. One of the most important series for comparison is the geometric series given by
\[1+x+x^{2}+\cdots.\]
Its properties are given below.
Proposition. Suppose that \(|x|<1\). Then, the geometric series \(1+x+x^{2}+\cdots\) is convergent given by
\begin{equation}{\label{maeq165}}\tag{26}
1+x+x^{2}+\cdots =\frac{1}{1-x}.
\end{equation}
Suppose that \(|x|\geq 1\). Then, the geometric series \(1+x+x^{2}+\cdots\) is divergent.
Proof. We have
\[(1-x)\sum_{k=0}^{n}x^{k}=\sum_{k=0}^{n}\left (x^{k}-x^{k+1}\right )=1-x^{n+1}.\]
Suppose that \(|x|<1\). Then, we have (\ref{maeq165}). Suppose that \(|x|\geq 1\). Then, the \(n\)th term \(x^{n}\) dose not tend to \(0\) as \(n\rightarrow\infty\), which says that the geometric series is divergent by Corollary \ref{mac152}. This completes the proof. \(\blacksquare\)
The concepts of limit superior and inferior can refer to the page Limit superior and inferior of sequences which will be used below.
\begin{equation}{\label{mat167}}\tag{27}\mbox{}\end{equation}
Theorem \ref{mat167}. (Ratio Test). Given a series \(\sum_{k=1}^{\infty}a_{k}\) with \({\color{blue}{a_{k}\neq 0}}\) for all \(k\), we define
\[r=\liminf_{n\rightarrow\infty}\left |\frac{a_{n+1}}{a_{n}}\right |\mbox{ and }R=\limsup_{n\rightarrow\infty}\left |\frac{a_{n+1}}{a_{n}}\right |.\]
Then, we have the following results.
(i) Suppose that \(R<1\). Then, the series \(\sum_{k=1}^{\infty}a_{k}\) is absolutely convergent.
(ii) Suppose that \(r>1\). Then, the series \(\sum_{k=1}^{\infty}a_{k}\) is divergent.
(iii) Suppose that \(r\leq 1\leq R\). Then, the test is inconclusive.
Proof. To prove part (i), we take \(x\) satisfying \(R<x<1\). By the definition of limit superior, we have
\begin{align*} R & =\limsup_{n\rightarrow\infty}\left |\frac{a_{n+1}}{a_{n}}\right |\\ & =\inf_{k\geq 1}
\sup_{n\geq k}\left |\frac{a_{n+1}}{a_{n}}\right |\\ & \equiv\inf_{k\geq 1}b_{k}=\lim_{k\rightarrow\infty}b_{k},\end{align*}
where the sequence \(\{b_{k}\}_{k=1}^{\infty}\) is decreasing given by
\[b_{k}=\sup_{n\geq k}\left |\frac{a_{n+1}}{a_{n}}\right |.\]
Since \(R<x<1\), there exists an integer \(N>0\) satisfying \(R<b_{N}\leq x<1\), i.e.,
\[b_{N}=\sup_{n\geq N}\left |\frac{a_{n+1}}{a_{n}}\right |\leq x.\]
Therefore, we obtain
\[\left |\frac{a_{n+1}}{a_{n}}\right |\leq x\mbox{ for }n\geq N.\]
Since \(x=x^{n+1}/x^{n}\), we have
\[\frac{|a_{n+1}|}{x^{n+1}}\leq\frac{|a_{n}|}{x^{n}}\leq\cdots\leq\frac{|a_{N}|}{x^{N}}\mbox{ for }n\geq N,\]
which implies
\[|a_{n}|\leq cx^{n}\mbox{ for }n\geq N,\mbox{ where }c=\frac{|a_{N}|}{x^{N}}.\]
Since \(x<1\), i.e., the geometric series \(\sum_{n=1}^{\infty}x^{n}\) is convergent, part (i) follows immediately from Theorem \ref{mat163}.
To prove part (ii), by the definition of limit inferior, we have
\begin{align*} r & =\liminf_{n\rightarrow\infty}\left |\frac{a_{n+1}}{a_{n}}\right |\\ & =\sup_{k\geq 1}
\inf_{n\geq k}\left |\frac{a_{n+1}}{a_{n}}\right |\\ & \equiv\sup_{k\geq 1}c_{k}=\lim_{k\rightarrow\infty}c_{k},\end{align*}
where the sequence \(\{c_{k}\}_{k=1}^{\infty}\) is increasing given by
\[c_{k}=\inf_{n\geq k}\left |\frac{a_{n+1}}{a_{n}}\right |.\]
Since \(r>1\), there exists an integer \(N>0\) satisfying \(r\geq c_{N}>1\). Therefore, we obtain \(|a_{n+1}|>|a_{n}|\) for all \(n\geq N\). This says that the sequence \(\{a_{n}\}_{n=1}^{\infty}\) cannot converge to \(0\), i.e., the series \(\sum_{k=1}^{\infty}a_{k}\) is divergent by Corollary \ref{mac152}. This proves part (ii).
To prove part (iii), we consider the series
\[\sum_{k=1}^{\infty}\frac{1}{k}\mbox{ and }\sum_{k=1}^{\infty}\frac{1}{k^{2}}.\]
In both cases, we have \(r=1=R\). However, the series \(\sum_{k=1}^{\infty}\frac{1}{k}\) is divergent, and the series \(\sum_{k=1}^{\infty}\frac{1}{k^{2}}\) is convergent. This completes the proof. \(\blacksquare\)
\begin{equation}{\label{mat168}}\tag{28}\mbox{}\end{equation}
Theorem \ref{mat168}. (Root Test). Given a series \(\sum_{k=1}^{\infty}a_{k}\), we define
\[\rho =\limsup_{n\rightarrow\infty}\sqrt[n]{|a_{n}|}.\]
Then, we have the following results.
(i) Suppose that \(\rho <1\). Then, the series \(\sum_{k=1}^{\infty}a_{k}\) is absolutely convergent.
(ii) Suppose that \(\rho >1\). Then, the series \(\sum_{k=1}^{\infty}a_{k}\) is divergent.
(iii) Suppose that \(\rho =1\). Then, the test is inconclusive.
Proof. Suppose that \(\rho <1\). We take \(x\) satisfying \(\rho <x<1\). According to the definition of limit superior, we have
\begin{*} \rho & =\limsup_{n\rightarrow\infty}\sqrt[n]{|a_{n}|}=\inf_{k\geq 1}\sup_{n\geq k}\sqrt[n]{|a_{n}|}\\ & \equiv\inf_{k\geq 1}b_{k}=
\lim_{k\rightarrow\infty}b_{k},\{*}
where the sequence \(\{b_{k}\}_{k=1}^{\infty}\) is decreasing given by
\[b_{k}=\sup_{n\geq k}\sqrt[n]{|a_{n}|}.\]
Since \(\rho <x<1\) and \(\lim_{k\rightarrow\infty}b_{k}=\rho\), there exists an integer \(N>0\) satisfying
\[\rho <b_{N}\leq x<1,\mbox{ i.e., }b_{N}=\sup_{n\geq N}\sqrt[n]{|a_{n}|}\leq x.\]
Therefore, we obtain \(|a_{n}|\leq x^{n}\) for all \(n\geq N\). Since \(x<1\), i.e., the geometric series \(\sum_{n=1}^{\infty}x^{n}\) is convergent, using Theorem \ref{mat163}, it follows that the series \(\sum_{k=1}^{\infty}|a_{k}|\) is convergent, which proves part (i). Suppose that \(\rho >1\), we have
\[b_{k}=\sup_{n\geq k}\sqrt[n]{|a_{n}|}>1\mbox{ for all }k\geq 1.\]
Therefore, for each \(k\geq 1\), there exists \(n_{k}\geq k\) satisfying \(\sqrt[n_{k}]{|a_{n_{k}}|}>1\), i.e., \(|a_{n_{k}}|>1\), which says that \(|a_{n}|>1\) are
infinitely often. Therefore, the sequence \(\{a_{n}\}_{n=1}^{\infty}\) cannot converge to \(0\), i.e., the series \(\sum_{k=1}^{\infty}a_{k}\) is divergent. This proves part (ii). To prove part (iii), the same examples in the proof of Theorem \ref{mat167} are still available in this case. This completes the proof. \(\blacksquare\)
\begin{equation}{\label{mat166}}\tag{29}\mbox{}\end{equation}
Theorem \ref{mat166}. (Integral Test). Let \(f\) be a decreasing function defined on \([1,+\infty )\) satisfying \(f(x)\geq M\) for some constant \(M\) and for all \(x\in [1,+\infty )\). We define
\[s_{n}=\sum_{k=1}^{n}f(k),\quad t_{n}=\int_{1}^{n}f(x)dx\mbox{ and }d_{n}=s_{n}-t_{n}.\]
Then, we have the following results.
(i) The following inequalities hold true
\[M\leq f(n+1)\leq d_{n+1}\leq d_{n}\leq f(1)\mbox{ for all }n.\]
This says that the sequence \(\{s_{n}\}_{n=1}^{\infty}\) is bounded, and the sequence \(\{d_{n}\}_{n=1}^{\infty}\) is decreasing and convergent with limit
\[\lim_{n\rightarrow\infty}d_{n}=\inf_{n\geq 1}d_{n}.\]
(ii) The series \(\sum_{k=1}^{\infty}f(k)\) is convergent if and only if the sequence \(\{t_{n}\}_{n=1}^{\infty}\) is convergent.
(iii) Suppose that \(f\) is nonnegative. Then, we have
\[0\leq d_{n}\leq f(n)+\lim_{k\rightarrow\infty}d_{k}\mbox{ for }n=1,2,\cdots .\]
Proof. To prove part (i), we have
\begin{align*}
t_{n+1} & =\int_{1}^{n+1}f(x)dx=\sum_{k=1}^{n}\int_{k}^{k+1}f(x)dx\leq
\sum_{k=1}^{n}\int_{k}^{k+1}f(k)dx\mbox{ (since \(f\) is decreasing)}\\
& =\sum_{k=1}^{n}f(k)=s_{n},
\end{align*}
which implies
\[f(n+1)=s_{n+1}-s_{n}\leq s_{n+1}-t_{n+1}=d_{n+1}.\]
On the other hand, we also have
\begin{align*}
d_{n}-d_{n+1} & =t_{n+1}-t_{n}-(s_{n+1}-s_{n})=\int_{n}^{n+1}f(x)dx-f(n+1)\\
& \geq\int_{n}^{n+1}f(n+1)dx-f(n+1)\mbox{ (since \(f\) is decreasing)}\\
& =0,
\end{align*}
which implies \(d_{n+1}\leq d_{n}\leq d_{1}=f(1)\). Using Proposition \ref{map150}, it follows that \(\{d_{n}\}_{n=1}^{\infty}\) is a bounded and decreasing sequence. Since \(s_{n}=d_{n}+t_{n}\) and the sequence \(\{d_{n}\}_{n=1}^{\infty}\) is convergent by part (i), part (ii) follows immediately by taking \(n\rightarrow\infty\).
To prove part (iii), for \(k>n\), we have
\begin{align*}
d_{k}-d_{n} & =\left (s_{k}-s_{n}\right )-\left (t_{k}-t_{n}\right )=\sum_{m=n+1}^{k}f(m)-\int_{n}^{k}f(x)dx\\
& =\sum_{m=n+1}^{k}f(m)-\sum_{m=n}^{k-1}\int_{m}^{m+1}f(x)dx\\
& \geq\sum_{m=n+1}^{k}f(m)-\sum_{m=n}^{k-1}\int_{m}^{m+1}f(m)dx\mbox{ (since \(f\) is decreasing)}\\
& =\sum_{m=n+1}^{k}f(m)-\sum_{m=n}^{k-1}f(m)=f(k)-f(n),
\end{align*}
which implies
\[d_{n}-d_{k}\leq f(n)-f(k)\leq f(n)\mbox{ (by the nonnegativity of \(f\))}\]
for all \(k>n\). By taking \(k\rightarrow\infty\) on both sides, we can obtain the desired inequalities. This completes the proof. \(\blacksquare\)
In Theorem \ref{mat166}, let
\[D=\lim_{n\rightarrow\infty}d_{n}.\]
Part (i) says \(0\leq D\leq f(1)\) and part (iii) says
\[0\leq\sum_{k=1}^{n}f(x)-\int_{1}^{n}f(x)dx\leq D+f(n),\]
which can be used to approximate the certain finite sums by integrals.
Proposition. Given two sequences \(\{a_{n}\}_{n=1}^{\infty}\) and \(\{b_{n}\}_{n=1}^{\infty}\), the \(n\)th partial sums of \(\{a_{n}\}_{n=1}^{\infty}\) is given by \(s_{n}=a_{1}+\cdots +a_{n}\). Then, we have the following identity
\begin{equation}{\label{ma22}}\tag{30}
\sum_{k=1}^{n}a_{k}b_{k}=s_{n}b_{n+1}-\sum_{k=1}^{n}s_{k}\left (b_{k+1}-b_{k}\right ).
\end{equation}
Therefore, if the series \(\sum_{k=1}^{\infty}s_{k}(b_{k+1}-b_{k})\) and the sequence \(\{s_{n}b_{n+1}\}_{n=1}^{\infty}\) are convergent, then the series \(\sum_{k=1}^{\infty}a_{k}b_{k}\) is convergent.
Proof. Let \(s_{0}=0\). Then, we have
\begin{align*} \sum_{k=1}^{n}a_{k}b_{k} & =\sum_{k=1}^{n}(s_{k}-s_{k-1})b_{k}\\ & =\sum_{k=1}^{n}s_{k}b_{k}
-\sum_{k=1}^{n}s_{k}b_{k+1}+s_{n}b_{n+1}\\ & =\sum_{k=1}^{n}s_{k}\left (b_{k}-b_{k+1}\right )+s_{n}b_{n+1}.\end{align*}
This completes the proof. \(\blacksquare\)
Theorem. (Dirichlets Test). Let \(\sum_{k=1}^{\infty}a_{k}\) be a series such that its partial sums form a bounded sequence. Suppose that \(\{b_{n}\}_{n=1}^{\infty}\) is a decreasing sequence that converges to \(0\). Then, the series \(\sum_{k=1}^{\infty}a_{k}b_{k}\) is convergent.
Proof. Let \(s_{n}\) be a partial sum of \(\sum_{k=1}^{\infty}a_{k}\). We assume that \(|s_{n}|\leq M\) for all \(n\). Then, we have
\begin{equation}{\label{ma121}}\tag{31}
\lim_{n\rightarrow\infty}|s_{n}b_{n+1}|\leq M\cdot\lim_{n\rightarrow\infty}b_{n+1}=0.
\end{equation}
Since the sequence \(\{b_{n}\}_{n=1}^{\infty}\) is decreasing, it follows
\[\left |s_{k}(b_{k+1}-b_{k})\right |\leq M\cdot(b_{k}-b_{k+1}).\]
Since the series \(\sum_{k=1}^{\infty}(b_{k+1}-b_{k})\) is convergent by Proposition \ref{map266}, the comparison test says that the series \(\sum_{k=1}^{\infty}s_{k}(b_{k+1}-b_{k})\) is absolutely convergent. Using (\ref{ma22}) and (\ref{ma121}), the proof is complete. \(\blacksquare\)
\begin{equation}{\label{mat169}}\tag{32}\mbox{}\end{equation}
Theorem \ref{mat169}. (Abel’s Test). Suppose that the series \(\sum_{k=1}^{\infty}a_{k}\) is convergent, and that \(\{b_{n}\}_{n=1}^{\infty}\) is a monotonic convergent sequence. Then, the series \(\sum_{k=1}^{\infty}a_{k}b_{k}\) is convergent.
Proof. Since the series \(\sum_{k=1}^{\infty}a_{k}\) is convergent, the sequence of partial sums \(\{s_{n}\}_{n=1}^{\infty}\) is bounded according to Proposition \ref{map162}. We assume that \(|s_{n}|\leq M\) for all \(n\). Since the sequence \(\{b_{n}\}_{n=1}^{\infty}\) is a monotonic convergent sequence, the limit
\[\lim_{n\rightarrow\infty}s_{n}b_{n+1}=\left (\lim_{n\rightarrow\infty}s_{n}\right ) \cdot\left (\lim_{n\rightarrow\infty}b_{n+1}\right ).\]
exists. From Proposition \ref{map266}, since the limit of the sequence \(\{b_{n}\}_{n=1}^{\infty}\) exists, the series \(\sum_{k=1}^{\infty}(b_{k+1}-b_{k})\) is convergent, which also says that the series \(\sum_{k=1}^{\infty}(b_{k}-b_{k+1})\) is convergent. Now, we have
\[\left |s_{k}(b_{k+1}-b_{k})\right |\leq\left\{\begin{array}{ll}
M\cdot (b_{k+1}-b_{k}) & \mbox{if \(\{b_{n}\}_{n=1}^{\infty}\) is increasing}\\
M\cdot (b_{k}-b_{k+1}) & \mbox{if \(\{b_{n}\}_{n=1}^{\infty}\) is decreasing}.
\end{array}\right .\]
Theorem \ref{mat163} says that the series \(\sum_{k=1}^{\infty}s_{k}(b_{k+1}-b_{k})\) is absolutely convergent. Using (\ref{ma22}), we also have
\[\sum_{k=1}^{n}s_{k}(b_{k+1}-b_{k})=s_{n}b_{n+1}-\sum_{k=1}^{n}a_{k}b_{k}.\]
By taking \(n\rightarrow\infty\), we see that the series \(\sum_{k=1}^{\infty}a_{k}b_{k}\) is convergent. This completes the proof. \(\blacksquare\)
\begin{equation}{\label{e}}\tag{E}\mbox{}\end{equation}
Double Sequence and Series.
A function \(f\) defined on \(\mathbb{N}\times\mathbb{N}\) is called a double sequence. We say that the double sequence \(f\) converges to \(a\) when, given any \(\epsilon >0\), there exists an integer \(N\) such that \(p>N\) and \(q>N\) imply \(|f(p,q)-a|<\epsilon\). This is denoted by
\[\lim_{p,q\rightarrow\infty}f(p,q)=a.\]
\begin{equation}{\label{ma120}}\tag{33}\mbox{}\end{equation}
Proposition \ref{ma120}. Suppose that
\[\lim_{p,q\rightarrow\infty}f(p,q)=a.\]
For each fixed \(p\), we assume that the limit
\[\lim_{q\rightarrow\infty}f(p,q)\]
exists. Then, we have
\[\lim_{p\rightarrow\infty}\left [\lim_{q\rightarrow\infty}f(p,q)\right ]=a=\lim_{p,q\rightarrow\infty}f(p,q).\]
A similar result still holds true if we interchange the roles of \(p\) and \(q\).
Proof. By the definition of double sequence, given any \(\epsilon >0\), there exits an integer \(N_{1}>0\) such that
\begin{equation}{\label{maeq267}}\tag{34}
p>N_{1}\mbox{ and }q>N_{1}\mbox{ imply }\left |f(p,q)-a\right |<\frac{\epsilon}{2}
\end{equation}
We define
\[F(p)=\lim_{q\rightarrow\infty}f(p,q).\]
Therefore, there exists an integer \(N_{2}>0\) such that
\begin{equation}{\label{maeq268}}\tag{35}
q>N_{2}\mbox{ implies }\left |F(p)-f(p,q)\right |<\frac{\epsilon}{2},
\end{equation}
where \(N_{2}\) depends on \(p\) and \(\epsilon\). Let
\[N=\max\{N_{1},N_{2}\}.\]
For \(p>N\) and \(q>N\), both (\ref{maeq267}) and (\ref{maeq268}) hold satisfying
\[\left |F(p)-a\right |\leq\left |F(p)-f(p,q)\right |+\left |f(p,q)-a\right |<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon ,\]
which says that \(F(p)\rightarrow a\) as \(p\rightarrow\infty\). This completes the proof. \(\blacksquare\)
To distinguish these two limits in Proposition \ref{ma120}, we call
\[\lim_{p,q\rightarrow\infty}f(p,q)=a\]
the double limit and
\[\lim_{p\rightarrow\infty}\left [\lim_{q\rightarrow\infty}f(p,q)\right ]\]
the iterated limit. Proposition \ref{ma120} says that the existence of double limit and of limit \(\lim_{q\rightarrow\infty}f(p,q)\) for each \(p\) implies the existence of iterated limit. However, the converse is not true, which can be realized in the following example.
Example. We consider the following double sequence
\[f(p,q)=\frac{pq}{p^{2}+q^{2}}.\]
Then, for each fixed \(p\), we have
\[\lim_{q\rightarrow\infty}f(p,q)=0=\lim_{p\rightarrow\infty}\left [\lim_{q\rightarrow\infty}f(p,q)\right ].\]
Since \(f(p,q)=1/2\) when \(p=q\) and \(f(p,q)=2/3\) when \(p=2q\), it says that the double limit does not exist. \(\sharp\)
Proposition. Let \(f\) be a double sequence, and let \(\mathbb{N}\) denote the set of positive integers. We define a function \(g_{n}\) on \(\mathbb{N}\) as \(g_{n}(m)=f(m,n)\) for \(m\in\mathbb{N}\). Suppose that \(g_{n}\rightarrow g\) uniformly on \(\mathbb{N}\), where
\begin{equation}{\label{maeq279}}\tag{36}
g(m)=\lim_{n\rightarrow\infty}f(m,n),
\end{equation}
and that the iterated limit
\[\lim_{m\rightarrow\infty}\left [\lim_{n\rightarrow\infty}f(m,n)\right ]\]
exists. Then, we have the double limit
\[\lim_{m,n\rightarrow\infty}f(m,n)=\lim_{m\rightarrow\infty}\left [\lim_{n\rightarrow\infty}f(m,n)\right ].\]
Proof. From (\ref{maeq279}), given any \(\epsilon >0\), there exists an integer \(N_{1}\) such that
\[n>N_{1}\mbox{ implies }\left |f(m,n)-g(m)\right |<\frac{\epsilon}{2}\mbox{ for all }m\in\mathbb{N}.\]
Let
\[a=\lim_{m\rightarrow\infty}\left [\lim_{n\rightarrow\infty}f(m,n)\right ]=\lim_{m\rightarrow\infty}g(m).\]
For the same \(\epsilon\), there exists an integer \(N_{2}>0\) such that
\[m>N_{2}\mbox{ implies }|g(m)-a|<\frac{\epsilon}{2}.\]
We take \(N=\max\{N_{1},N_{2}\}\). Then \(m>N\) and \(n>N\) implies
\[\left |f(m,n)-a\right |\leq\left |f(m,n)-g(m)\right |+|g(m)-a|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.\]
This shows
\[\lim_{m,n\rightarrow\infty}f(m,n)=a,\]
and the proof is complete. \(\blacksquare\)
Definition. Let \(f\) be a double sequence, and let \(s\) be the double sequence defined by
\[s(p,q)=\sum_{m=1}^{p}\sum_{n=1}^{q}f(m,n).\]
The pair \((f,s)\) is called a {\bf double series} and is denoted by
\[\sum_{m,n=1}^{\infty}f(m,n).\]
The double series is said to converge to the sum \(a\) when
\[\lim_{p,q\rightarrow\infty}s(p,q)=a.\]
The term \(s(p,q)\) is also called the partial sum of the double series. We say that the double series \(\sum_{m,n=1}^{\infty}f(m,n)\) is absolutely convergence when the double series \(\sum_{m,n=1}^{\infty}|f(m,n)|\) is convergent. \(\sharp\)
Proposition. We have the following properties.
(i) A double series of positive terms is convergent if and only if the set of partial sums is bounded.
(ii) A double series is convergent if it is absolutely convergent. \(\sharp\)
Let \(f\) be a double sequence, and let \(\eta :\mathbb{N}\rightarrow\mathbb{N}\times\mathbb{N}\) be a one-to-one function. Suppose that \(g\) is the sequence defined by
\[g(n)=f(\eta (n))\mbox{ for }n\in\mathbb{N}.\]
Then \(\eta\) is said to be an arrangement of the double sequence \(f\) into the sequence \(g\).
\begin{equation}{\label{map269}}\tag{37}\mbox{}\end{equation}
Proposition \ref{map269}. Let \(\sum_{m,n=1}^{\infty}f(m,n)\) be a double series, and let \(\eta\) be an arrangement of the double sequence \(f\) into a sequence \(g\). Then, we have the following properties.
(i) The series \(\sum_{n=1}^{\infty}g(n)\) is absolutely convergent if and only if the double series \(\sum_{m,n=1}^{\infty}f(m,n)\) is absolutely convergent.
(ii) Suppose that the double series \(\sum_{m,n=1}^{\infty}f(m,n)\) is absolutely convergent with sum \(s\). Then, we have the following properties.
- We have \(\sum_{n=1}^{\infty}g(n)=s\).
- Both series \(\sum_{n=1}^{\infty}f(m_{0},n)\) and \(\sum_{m=1}^{\infty}f(m,n_{0})\) are absolutely convergent.
- Define
\[A_{m_{0}}=\sum_{n=1}^{\infty}f(m_{0},n)\mbox{ and }B_{n_{0}}=\sum_{m=1}^{\infty}f(m,n_{0}).\]
Then, both series \(\sum_{m=1}^{\infty}A_{m}\) and \(\sum_{n=1}^{\infty}B_{n}\) are absolutely convergent and have the same sum \(s\); that is,
\begin{equation}{\label{maeq172}}\tag{38}
\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}f(m,n)=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}f(m,n)=s.
\end{equation}
The series shown in (\ref{maeq172}) are called iterated series. The convergence of both iterated series does not imply their equality. The example is given below. We consider the double sequence
\[f(m,n)=\left\{\begin{array}{ll}
1 & \mbox{if \(m=n+1\) for \(n=1,2,\cdots\)}\\
-1 & \mbox{if \(m=n-1\) for \(n=1,2,\cdots\)}\\
0 & \mbox{otherwise}.
\end{array}\right .\]
Then, we have
\[\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}f(m,n)=-1\mbox{ and }\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}f(m,n)=1.\]
The sufficient conditions for guaranteeing the equality of those two iterated series are given below.
\begin{equation}{\label{mat270}}\tag{39}\mbox{}\end{equation}
Theorem \ref{mat270}. Let \(f\) be a double sequence. Suppose that the series \(\sum_{n=1}^{\infty}f(m_{0},n)\) is absolutely convergent for each fixed \(m_{0}\), and that the iterated series \(\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}|f(m,n)|\) is convergent.
Then, we have the following properties.
(i) The double series \(\sum_{m,n=1}^{\infty}f(m,n)\) is absolutely convergent.
(ii) The series \(\sum_{m=1}^{\infty}f(m,n_{0})\) is absolutely convergent for each fixed \(n_{0}\).
(iii) We have
\[\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}f(m,n)=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}f(m,n)=\sum_{m,n=1}^{\infty}f(m,n).\]
Proof. Let \(\eta\) be an arrangement of the double sequence \(f\) into the sequence \(g\). Since all the partial sums of the series \(\sum_{k}|g(k)|\) are bounded by the convergence of \(\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}|f(m,n)|\), the series \(\sum_{k}g(k)\) is absolutely convergent. Part (i) of Proposition \ref{map269} says that the double series \(\sum_{m,n=1}^{\infty}f(m,n)\) is absolutely convergent, which proves part (i). Parts (ii) and (iii) also follow from Proposition \ref{map269}, and the proof is complete. \(\blacksquare\)
\begin{equation}{\label{map173}}\tag{40}\mbox{}\end{equation}
Proposition \ref{map173}. Suppose that both series \(\sum_{m=1}^{\infty}a_{m}\) and \(\sum_{n=1}^{\infty}b_{n}\) are absolutely convergent with sums \(A\) and \(B\), respectively. Let \(f\) be a double sequence defined by \(f(m,n)=a_{m}b_{n}\). Then, the double series
\[\sum_{m,n=1}^{\infty}f(m,n)=\sum_{m,n=1}a_{m}b_{n}\]
is absolutely convergent and has the sum \(A\cdot B\).
Proof. We have
\begin{align*} \left (\sum_{m=1}^{\infty}\left |a_{m}\right |\right )\left (\sum_{n=1}^{\infty}\left |b_{n}\right |\right ) & =\sum_{m=1}^{\infty}\left (\left |a_{m}\right |\sum_{n=1}^{\infty}\left |b_{n}\right |\right )\\ & =\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}
\left |a_{m}\right |\left |b_{n}\right |.\end{align*}
Theorem \ref{mat270} says that the double series \(\sum_{m,n=1}a_{m}b_{n}\) is absolutely convergent and has the sum \(AB\). This completes the proof. \(\blacksquare\)
Given two absolutely convergent power series
\[\sum_{k=0}^{\infty}a_{k}x^{k}\mbox{ and }\sum_{k=0}^{\infty}b_{k}x^{k}\]
with sums \(A(x)\) and \(B(x)\), respectively. We can show
\[A(x)\cdot B(x)=\sum_{k=0}^{\infty}c_{k}x^{k},\]
where
\begin{equation}{\label{maeq271}}\tag{41}
c_{n}=\sum_{k=0}^{n}a_{k}b_{n-k}.
\end{equation}
In general, we define the so-called Cauchy product below.
Definition. Given two series
\[\sum_{k=0}^{\infty}a_{k}\mbox{ and }\sum_{k=0}^{\infty}b_{k},\]
we define
\[c_{n}=\sum_{k=0}^{n}a_{k}b_{n-k}.\]
The series \(\sum_{k=0}^{\infty}c_{k}\) is called the Cauchy product of \(\sum_{k=0}^{\infty}a_{k}\) and \(\sum_{k=0}^{\infty}b_{k}\). \(\sharp\)
Suppose that both series
\[\sum_{k=0}^{\infty}a_{k}\mbox{ and }\sum_{k=0}^{\infty}b_{k}\]
are absolutely convergent. Then, using Propositions \ref{map173} and \ref{map155}, the Cauchy product converges to the following value
\begin{equation}{\label{maeq174}}\tag{42}
\sum_{k=0}^{\infty}c_{k}=\left (\sum_{k=0}^{\infty}a_{k}\right )\left (\sum_{k=0}^{\infty}b_{k}\right ).
\end{equation}
The equality (\ref{maeq174}) may fail to hold when both series \(\sum_{k=0}^{\infty}a_{k}\) and \(\sum_{k=0}^{\infty}b_{k}\) are conditionally convergent. However, we can prove that equality (\ref{maeq174}) is valid when at least one of \(\sum_{k=0}^{\infty}a_{k}\) and \(\sum_{k=0}^{\infty}b_{k}\) is absolutely convergent.
\begin{equation}{\label{mat293}}\tag{43}\mbox{}\end{equation}
Theorem \ref{mat293}. Suppose that the series \(\sum_{k=0}^{\infty}a_{k}\) is absolutely convergent with sum \(A\), and that the series \(\sum_{k=0}^{\infty}b_{k}\) is convergent with sum \(B\). Then, the Cauchy product of these two series \(\sum_{k=0}^{\infty}a_{k}\) and \(\sum_{k=0}^{\infty}b_{k}\) are convergent with sum \(A\cdot B\).
Proof. Let \(c_{k}\) be defined in (\ref{maeq271}). We define the following partial sums
\[A_{n}=\sum_{k=0}^{n}a_{k},\quad B_{n}=\sum_{k=0}^{n}b_{k}\mbox{ and }C_{n}=\sum_{k=0}^{n}c_{k}.\]
Let
\[d_{n}=B-B_{n}\mbox{ and }e_{n}=\sum_{k=0}^{n}a_{k}d_{n-k}.\]
Then, we have
\begin{align*}
C_{p} & =\sum_{n=0}^{p}\sum_{k=0}^{n}a_{k}b_{n-k}=\sum_{k=0}^{p}\sum_{n=k}^{p}a_{k}b_{n-k}=\sum_{k=0}^{p}a_{k}
\left (\sum_{m=0}^{p-k}b_{m}\right )\\
& =\sum_{k=0}^{p}a_{k}B_{p-k}=\sum_{k=0}^{p}a_{k}(B-d_{p-k})=A_{p}B-e_{p}.
\end{align*}
It is obvious that the sequence \(\{d_{n}\}_{n=1}^{\infty}\) is bounded and converges to \(0\). We can take \(M>0\) satisfying \(|d_{n}|\leq M\) for all \(n\). Let \(K=\sum_{n=0}^{\infty}|a_{n}|\). Since \(\{d_{n}\}_{n=1}^{\infty}\) converges to \(0\), given any \(\epsilon >0\), there exists \(N>0\) such that \(n>N\) implies \(|d_{n}|<\epsilon /(2K)\). The convergence of \(\sum_{n=0}^{\infty}|a_{n}|\) also says
\[\sum_{n=N+1}^{\infty}\left |a_{n}\right |<\frac{\epsilon}{2M}.\]
Therefore, for \(p>2N\), we have
\begin{align*}
\left |e_{p}\right | & =\left |\sum_{k=0}^{p}a_{k}d_{p-k}\right |
\leq\sum_{k=0}^{N}\left |a_{k}d_{p-k}\right |+\sum_{k=N+1}^{p}\left |a_{k}d_{p-k}\right |\\
& \leq\frac{\epsilon}{2K}\sum_{k=0}^{N}\left |a_{k}\right |+M\sum_{k=N+1}^{p}\left |a_{k}\right |\\
& \leq\frac{\epsilon}{2K}\sum_{k=0}^{\infty}\left |a_{k}\right |+
M\sum_{k=N+1}^{\infty}\left |a_{k}\right |<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon .
\end{align*}
This shows that \(e_{p}\rightarrow 0\) as \(p\rightarrow\infty\). Therefore, we obtain
\begin{align*} \sum_{k=0}^{\infty}c_{k} & =\lim_{p\rightarrow\infty}C_{p}\\ & =\lim_{p\rightarrow\infty}
\left (A_{p}B-e_{p}\right )\\ & =B\cdot\lim_{p\rightarrow\infty}A_{p}-\lim_{p\rightarrow\infty}e_{p}=AB.\end{align*}
This completes the proof. \(\blacksquare\)
