This problem still poses some challenge. Since $a_n$ is an alternating term, I wouldn't know I to go about it. The question is, if $a_n>0,a_{n+1}<a_n, \;a_n\to 0.$ Define $$x_1=a_1\;\; \text{and}\;\; x_{n+1}=x_n+(-1)^n a_{n+1}.\;\;\;\;(*)$$
Is $\{x_n\}$ convergent? Below is a useful hint:
Since $\{x_n\}$ is a sequence in $\Bbb{R}$ and $\Bbb{R}$ is complete, it is sufficient to prove that $\Bbb{R}$ is Cauchy. Also, I am required to use the conditions $a_n>0,a_{n+1}<a_n, \;a_n\to 0$ and the term $(-1)^n a_{n+1}$. So, rewriting $(*)$ we get:
$$\sum_{n=1}^{\infty}(x_{n+1}-x_n)=\sum_{n=1}^{\infty}(-1)^n a_{n+1}. \;\;\;\;(**)$$
I want to use the alternating series test to obtain that $\sum_{n=1}^{\infty}(x_{n+1}-x_n)$ converges. Also, letting $m,n\in\Bbb{N}$ with $n>m,$ I am to use $(**)$ to prove that $|x_n-x_m|\to 0\;\;\text{as}\;\;n\to \infty.$
$x_n =x_1 +\sum_{k=2}^n (x_k -x_{k-1} ) =a_1 + \sum_{k=2}^n (-1)^{k-1} a_k =\sum_{k=1}^n (-1)^{k-1} a_k\to \sum_{k=1}^{\infty} (-1)^{k-1} a_k$ by Leibnitz Test.