Show that $\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\frac{1}{n^2\log(1+\frac{k^2}{n^2})}=\frac{{\pi}^2}{6}$ How can I expand this following limit?
$$\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{n^2\log(1+\frac{k^2}{n^2})}=\frac{{\pi}^2}{6}.$$
 A: For every $x$ in $[0,1]$, $1/(1+x)^2\leqslant1/(1+x)\leqslant 1$ hence, integrating from $0$ to $x$, 
$$
\frac{x}{1+x}\leqslant\log(1+x)\leqslant x.
$$
Applying this to every $x=k^{\color{red}{\alpha}}/n^\color{red}{\alpha}$ for some positive $\color{red}{\alpha}$, one sees that the sum 
$$
S_n^{(\color{red}{\alpha})}=\sum_{k=1}^n\frac1{n^\color{red}{\alpha}\log\left(1+\frac{k^{\color{red}{\alpha}}}{n^\color{red}{\alpha}}\right)}
$$ 
is such that
$$
\sum_{k=1}^n\frac1{n^{\color{red}{\alpha}}}\frac1{\frac{k^{\color{red}{\alpha}}}{n^{\color{red}{\alpha}}}}\leqslant S_n^{(\color{red}{\alpha})}\leqslant\sum_{k=1}^n\frac1{n^\color{red}{\alpha}}\frac{1+\frac{k^\color{red}{\alpha}}{n^\color{red}{\alpha}}}{\frac{k^\color{red}{\alpha}}{n^\color{red}{\alpha}}},
$$
that is,
$$
\sum_{k=1}^n\frac1{k^\color{red}{\alpha}}\leqslant S_n^{(\color{red}{\alpha})}\leqslant\frac1{n^{\color{red}{\alpha}-1}}+\sum_{k=1}^n\frac1{k^\color{red}{\alpha}}.
$$
In particular, $S_n^{(\color{red}{\alpha})}\to\infty$ when $\color{red}{\alpha}\leqslant1$ and, for every $\color{red}{\alpha}\gt1$,
$$
\lim_{n\to\infty}S_n^{(\color{red}{\alpha})}=\sum_{k=1}^\infty\frac1{k^\color{red}{\alpha}}=\zeta(\color{red}{\alpha}).
$$
