# How can I solve <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-TeX

How can I solve
$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{n}^{2}}\sum _{k=0}^{n}\frac{1}{\mathrm{ln}\left(1+\frac{\left(n+k\right)\sqrt{{n}^{2}+{k}^{2}}}{{n}^{3}}\right)}$
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iberistazi
$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{n}^{2}}\sum _{k=0}^{n}\frac{1}{\mathrm{log}\left(1+\frac{\left(n+k\right)\sqrt{{n}^{2}+{k}^{2}}}{{n}^{3}}\right)}=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{n}^{2}}\sum _{k=0}^{n}\frac{1}{\mathrm{log}\left(1+\frac{1}{n}\left(1+\frac{k}{n}\right)\sqrt{1+\frac{{k}^{2}}{{n}^{2}}}\right)}.$
Now, remembering that $x\ge \mathrm{log}\left(1+x\right)$ for every $x>0$ and that for every $\epsilon >0$ there is a $\delta >0$ such that for every $x\in \left(0,\delta \right)$ we have $\mathrm{log}\left(1+x\right)\ge \left(1-\epsilon \right)x$
The first inequality implies
$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{n}^{2}}\sum _{k=0}^{n}\frac{1}{\mathrm{log}\left(1+\frac{1}{n}\left(1+\frac{k}{n}\right)\sqrt{1+\frac{{k}^{2}}{{n}^{2}}}\right)}\ge \underset{n\to \mathrm{\infty }}{lim}\frac{1}{{n}^{2}}\sum _{k=0}^{n}\frac{1}{\frac{1}{n}\left(1+\frac{k}{n}\right)\sqrt{1+\frac{{k}^{2}}{{n}^{2}}}}=$
$=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}\sum _{k=0}^{n}\frac{1}{\left(1+\frac{k}{n}\right)\sqrt{1+\frac{{k}^{2}}{{n}^{2}}}}={\int }_{0}^{1}\frac{1}{\left(1+x\right)\sqrt{1+{x}^{2}}}dx.$
On the other hand, since (for a fixed $\epsilon$) for n big enough we have that for every k it holds that $\frac{1}{n}\left(1+\frac{k}{n}\right)\sqrt{1+\frac{{k}^{2}}{{n}^{2}}}\le \frac{2\sqrt{2}}{n}\le \delta$
$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{n}^{2}}\sum _{k=0}^{n}\frac{1}{\mathrm{log}\left(1+\frac{1}{n}\left(1+\frac{k}{n}\right)\sqrt{1+\frac{{k}^{2}}{{n}^{2}}}\right)}\le \frac{1}{1-\epsilon }{\int }_{0}^{1}\frac{1}{\left(1+x\right)\sqrt{1+{x}^{2}}}dx.$
Since $\epsilon$ is arbitrary, we get
$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{n}^{2}}\sum _{k=0}^{n}\frac{1}{\mathrm{log}\left(1+\frac{1}{n}\left(1+\frac{k}{n}\right)\sqrt{1+\frac{{k}^{2}}{{n}^{2}}}\right)}={\int }_{0}^{1}\frac{1}{\left(1+x\right)\sqrt{1+{x}^{2}}}dx.$