TITLE: A limit involving $\frac{\pi^2}{6}$
QUESTION [8 upvotes]: Let $a_n=1+\frac{1}{2^2}+\dots+\frac{1}{n^2}$. Considering known that $\lim_{n\to\infty}a_n=\frac{\pi^2}{6}$, evaluate $$\lim_{n\to\infty}n\left( a_n-\frac{\pi^2}{6}\right)$$
My attempt: I first proved that $b_n=n\left( a_n-\frac{\pi^2}{6}\right)$ is decreasing , i.e. $b_{n+1}-b_n=a_n+\frac{1}{n+1}-\frac{\pi^2}{6} \leq 0$, which is true since $$\lim_{n\to\infty}\left(a_n+\frac{1}{n+1}\right)=\frac{\pi^2}{6}$$ and $a_n+\frac{1}{n+1}$ is an increasing sequence.
Then I proved that $b_n$ is bounded by $0$ (obviously) and $-1$, the latter being true since it is equivalent to $a_n+\frac{1}{n}\geq \frac{\pi^2}{6}$, which can be proven as above. 
So $b_n$ is both decreasing and bounded, which means that it has a limit $l$ and I ended with Stolz-Cesaro:
$$l=\lim_{n\to\infty}n\left( a_n-\frac{\pi^2}{6}\right)=\lim_{n\to\infty}\frac{n^2\left( a_n-\frac{\pi^2}{6}\right)}{n}=\lim_{n\to\infty}\left( (n+1)^2\left( a_{n+1}-\frac{\pi^2}{6}\right) -n^2\left( a_n-\frac{\pi^2}{6}\right)\right)=\lim_{n\to\infty}\left( \frac{n^2}{(n+1)^2}+a_{n+1}-\frac{\pi^2}{6}+2n\left( a_{n+1}-\frac{\pi^2}{6}\right)\right)=1+0+2l$$ so $l=-1$.
Can anyone provide a shorter solution, if there is one, please?

REPLY [10 votes]: 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}