On reciprocal squares and infinite sums

Question

I have this problem here, but I am unable to find a proper solution for it. The problem is as follows:

For every positive integer $n$, we take $$a_n =\sum_{i=1}^{n} \frac{1}{i^2}$$ We have to prove that $$\sum_{n\geq 2} \frac{1}{n^2a_na_{n-1}}$$ converges and find its value. Any hints on the direction of approach is highly appreciated. Thank you.

Answer 1

Hint: $$\frac{1}{n^2a_n a_{n - 1}} = \frac{1}{n^2} \cdot \frac{1}{a_n a_{n - 1}} = (a_n - a_{n - 1}) \frac{1}{a_n a_{n - 1}} = \frac{1}{a_{n - 1}} - \frac{1}{a_n}$$

Answer 2

Convergence is easy since $a_n\ge1$ for all $n$. To find the value observe that $$ \frac{1}{a_n\,a_{n-1}}=n^2\Bigl(\frac{1}{a_{n-1}}-\frac{1}{a_n}\Bigr). $$