A limit involving $\frac{\pi^2}{6}$

Note that

$$\frac{\pi^2}6-a_n=\sum_{k=n+1}^\infty\frac1{k^2}$$

And

$$\frac1{n+1}=\int_{n+1}^\infty\frac1{x^2}~dx<\sum_{k=n+1}^\infty\frac1{k^2}<\int_n^\infty\frac1{x^2}~dx=\frac1n$$

Therefore, by the squeeze theorem, the limit is $-1$.

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If one wishes to avoid integrals, creative telescoping may be employed:

$$\frac1k-\frac1{k+1}=\frac1{k(k+1)}<\frac1{k^2}<\frac1{(k-1)k}=\frac1{k-1}-\frac1k$$

I leave the telescoping process to the reader. $\ddot\smile$

You may like this. By the Stolz-Cesaro Theorem, one has \begin{eqnarray} &&\lim_{n\to\infty}n\left( a_n-\frac{\pi^2}{6}\right)\\ &=&\lim_{n\to\infty}\frac{ a_n-\frac{\pi^2}{6}}{\frac1n}\\ &=&\lim_{n\to\infty}\frac{(a_{n+1}-\frac{\pi^2}{6})- (a_n-\frac{\pi^2}{6})}{\frac1{n+1}-\frac1n}\\ &=&\lim_{n\to\infty}\frac{a_{n+1}-a_n}{-\frac1{n(n+1)}}\\ &=&\lim_{n\to\infty}\frac{\frac1{(n+1)^2}}{-\frac1{n(n+1)}}\\ &=&-1. \end{eqnarray}