1
$\begingroup$

Let $\{a_n\}$ be a monotone decreasing sequence of positive real numbers converging to $0.$ I want to prove that $\sum^{\infty}_{n=1}(-1)^{n}a_n$ converges.

MY TRIAL

My idea is to use the sequence of partial sums and use monotone convergence criterion. Let \begin{align}\sum^{\infty}_{n=1}(-1)^{n}a_n=\lim\limits_{n\to\infty}s_n=\lim\limits_{n\to\infty}\sum^{n}_{k=1}(-1)^{k}a_k\end{align}

MONOTONICITY

For even terms, let $n=2m+1.$ Then, \begin{align}s_{2m+3}&=s_{2m+1}+a_{2m+2}-a_{2m+3}\\&=s_{2m+1}+(a_{2m+2}-a_{2m+3})\end{align} Since \begin{align}a_{n+1}\leq a_{n},\;\forall\;n\in\;\Bbb{N}\end{align} then \begin{align}a_{2m+2}-a_{2m+3}\geq 0\end{align} Therefore, \begin{align}s_{2m+3}\geq s_{2m+1},\;\forall\;m\in\;\Bbb{N}\end{align} So, $\{s_{2m+1}\}$ is a non-decreasing sequence.

BOUNDEDNESS

\begin{align}s_{2m+1}&=\sum^{2m+1}_{k=1}(-1)^{k}a_k\\&=-(a_1-a_2)-(a_3-a_4)-\cdots-(a_{2m-1}-a_{2m})-a_{2m+1}\\&\leq-a_{2m+1}\end{align} Since $\{s_{2m+1}\}$ is a non-decreasing sequence, bounded above. Then, by monotone convergence theorem, it converges. Let \begin{align}\lim\limits_{n\to\infty}s_{2m+3}=s\end{align} If $n$ is even, say $n=2m+2$, then the sum of the first $n$ terms is \begin{align}s_n\equiv s_{2m+2}=s_{2m+1}+a_{2m+2}\end{align} By hypothesis $a_n\to 0$, as $n\to \infty,$ so that \begin{align}\lim\limits_{n\to\infty}a_{2m+2}=0\end{align} and as $m\to \infty,$ \begin{align}s_n\equiv s_{2m+2}=s_{2m+1}+a_{2m+2}\to s\end{align} Therefore, \begin{align}\lim\limits_{n\to\infty}s_{n}=s\end{align}

Kindly check, I'm I correct? If no, corrections are highly welcome! If I am totally wrong, alternative proofs or counterexamples will be appreciated.

$\endgroup$
  • $\begingroup$ en.wikipedia.org/wiki/Alternating_series_test $\endgroup$ – Riemann Sep 6 '18 at 6:17
  • $\begingroup$ @Riemann: The series in en.wikipedia.org/wiki/Alternating_series_test, starts from $n=0$ but mine starts from $n=1.$ Kindly compare. Thanks! $\endgroup$ – Micheal Sep 6 '18 at 6:22
  • 1
    $\begingroup$ it is not essential distinct! $\endgroup$ – Riemann Sep 6 '18 at 6:23
  • 1
    $\begingroup$ Dear Downvoter, why the downvote? I wrote this proof myself! Check the content, they are very different from wikipedia or other posts! $\endgroup$ – Micheal Sep 6 '18 at 6:32
  • 1
    $\begingroup$ Do know $\sum^{\infty}_{n=1}(-1)^{n}a_n$ and $\sum^{\infty}_{n=0}(-1)^{n}a_n$ has the same convergence? $\endgroup$ – Riemann Sep 6 '18 at 6:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.