How can I expand this following limit? $$\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{n^2\log(1+\frac{k^2}{n^2})}=\frac{{\pi}^2}{6}.$$
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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 $\color{red}{\alpha}\gt1$, $$ \lim_{n\to\infty}S_n^{(\color{red}{\alpha})}=\sum_{k=1}^\infty\frac1{k^\color{red}{\alpha}}=\zeta(\color{red}{\alpha}). $$
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