<s>[INST] Limit of the sum using integral $\lim\limits_{n\rightarrow\infty}\sum_{k = 1}^{n} \frac{1}{(k+n)\sqrt{1 + n\ln({1+\frac{k}{n^2}})}}$. I can find it using integral: $\lim\limits_{n\rightarrow\infty}\frac{1}{n}\sum_{k = 1}^{n} \frac{1}{(1 + \frac{k}{n})\sqrt{1 + n\ln({1+\frac{k}{n^2}})}}$, but how get rid of $n$ and $\frac{k}{n^2}$ in sqrt? Gan you give me a hint, please?
 [/INST] We have
\begin{align}
\frac 1{\sqrt{1 + n\ln\left({1+\frac{k}{n^2}}\right)}}
&=\left(1 + n\ln\left({1+\frac{k}{n^2}}\right)\right)^{-1/2}\\
&=\left(1 + \frac kn+O\left(\frac{k^2}{n^3}\right)\right)^{-1/2}\\
&=\frac 1{\sqrt{1 + \frac kn}}+\frac 1nO\left(\frac{k}{n}\right)^2
\end{align}
Consequently,
\begin{align}
\sum_{k = 1}^{n} \frac{1}{(k+n)\sqrt{1 + n\ln({1+\frac{k}{n^2}})}}
&=\frac 1n\sum_{k = 1}^{n} \frac{1}{(1+\frac kn)\sqrt{1 + \frac kn}}+\frac 1n\sum_{k = 1}^{n} \frac{O\left(\frac{k^2}{n^3}\right)}{(1+\frac kn)\sqrt{1 + \frac kn}}\\
&=\frac 1n\sum_{k = 1}^{n}\left(1+\frac kn\right)^{-3/2}+\frac 1{n^2}\sum_{k = 1}^{n} O\left(\frac{k^2}{n^2}\right)\left(1+\frac kn\right)^{-3/2}
\end{align}
and the last sum vanish as $n\to\infty$.
</s>