# Closed form of an Infinite Series $\sum\limits_{k=1}^{\infty}\frac{1}{k^2a_ka_{k+1}}$

Let $a_n=\sum\limits_{k=1}^{n}\frac{1}{n^2}$ then define $S=\sum\limits_{k=1}^{\infty}\frac{1}{k^2a_ka_{k+1}}$. Question : Does the series converge?

Trivially we see $1<a_n<\frac{\pi^2}{6};\ \forall n\ge 1$. Thus $S\le \sum\frac{1}{k^2}=\frac{\pi^2}{6}$. Thus converges. But what about the closed form of the limit? I am stuck here.