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.