How to find $\lim_{n\to\infty}\left(\frac{\pi^2}{6}-\sum_{k=1}^n\frac{1}{k^2}\right)n$? How to find $\lim_{n\to\infty}\left(\frac{\pi^2}{6}-\sum_{k=1}^n\frac{1}{k^2}\right)n$?
It is well-known that $\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k^2}=\frac{\pi^2}{6}$, so $$\lim_{n\to\infty}\left(\frac{\pi^2}{6}-\sum_{k=1}^n\frac{1}{k^2}\right)n=\lim_{n\to\infty}\frac{\frac{\pi^2}{6}-\sum_{k=1}^n\frac{1}{k^2}}{\frac{1}{n}}$$ has indeterminate form $\frac{0}{0}$.
But it can't use L'Hôpital's Rule.
 A: $$n\sum_{k=n+1}^\infty \frac{1}{k^2}=\frac{1}{n}\sum_{k=n+1}^\infty\frac{1}{(k/n)^2}$$
This looks like a Riemann sum:
$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=n+1}^\infty\frac{1}{(k/n)^2}=\int_{1}^\infty\frac{1}{x^2}dx=1$$
where $\frac{1}{n}$ plays the role of $dx$ in the limit and $x=k/n$.
If worried about the infinite upper limit, it's ok - because the integral of $1/x^2$ is absolutely convergent, the upper limit can be replaced by $k_{max}=Mn$ ($M>1$) and limited separately after the limit in $n$, and shown that it converges for any $M$.
A: We may also use creative telescoping. Since: $$\left(\frac{1}{n}-\frac{1}{n+1}\right)\leq\frac{1}{n^2}\leq \left(\frac{1}{n-1}-\frac{1}{n}\right)\tag{1}$$
it follows that:
$$ \frac{1}{n+1}\leq\sum_{k>n}\frac{1}{k^2}\leq \frac{1}{n}, \tag{2}$$
hence the limit is $\color{red}{1}$ by squeezing.
A: Alternatively, you can use the Stolz-Cesaro theorem, the counterpart of L'Hospital's rule for sequences. It yields that your limit is the same as $$\lim_{n\to\infty}\frac{1/(n+1)^2}{1/n - 1/(n+1)}=1. $$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\lim_{n \to \infty}\bracks{\pars{{\pi^{2} \over 6} - \sum_{k = 1}^{n}{1 \over k^{2}}}n} &=
\lim_{n \to \infty}\bracks{\pars{{1 \over n} -
2\int_{n}^{\infty}{\braces{x} \over x^{3}}\,\dd x}n}\label{1}\tag{1}
\end{align}

where we used a
  well known identity which is related to the Riemann Zeta function.  

 Moreover,
$\ds{0 < 2\int_{n}^{\infty}{\braces{x} \over x^{3}}\,\dd x < {1 \over n^{2}}}$
such that \eqref{1} becomes:
$$\bbx{\ds{%
\lim_{n \to \infty}\bracks{\pars{{\pi^{2} \over 6} -
\sum_{k = 1}^{n}{1 \over k^{2}}}n} = 1}}
$$
