how to solve $\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}$? $\displaystyle\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+1}}\right)^{n}\ge\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+k}}\right)^{n}\ge\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+n}}\right)^{n}$
left=$\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+1}}}\right)^{n}}=e^{\displaystyle n \ln{\frac{n}{\sqrt{n^2+1}}} }=e^{0}=1$
right=$\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+n}}}\right)^{n}}=e^{\displaystyle n \ln{\frac{n}{\sqrt{n^2+n}}} }=e^{-\frac{1}{2}}$
left $\ne$ right ,what to do next?

$\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}=\lim _{n \rightarrow \infty} e^{\displaystyle n \ln \sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+k}}(1)}$
$(1)=\displaystyle \lim _{n \rightarrow \infty} n\left(\ln \frac{1}{n} \sum_{k=1}^{n} \frac{1}{\sqrt{1+k/n^{2}}}\right)$$=\lim _{n \rightarrow \infty} n\left(\ln \frac{1}{n} \sum_{k=1}^{n} \frac{1}{\sqrt{1+k/n \cdot 1/n}}\right)$$=\lim _{n \rightarrow \infty} n \ln \int_{0}^{1} \frac{1}{\sqrt{1+x/n}} d x$$=\lim _{n \rightarrow \infty} n \ln \int_{0}^{1} \frac{nd(x/n+1)}{\sqrt{1+x/n}}$$= \lim_{n\to\infty}{n\ln{n \cdot2 \left.\sqrt{1+\frac{x}{n} }\right|_{0}^{1}}}$$= \lim_{n\to\infty}{n\ln{n \cdot2 (\sqrt{1+\frac{1}{n}}-1)}}$$=\lim_{n\to\infty}{n\ln{n \cdot2 (\frac{1}{2n} -\frac{1}{2n^2} +o(\frac{1}{n^2}))}}$$=\lim_{n\to\infty}{n\ln{n \cdot2 (\frac{1}{2n} +\left(\frac{1}{2!}\cdot \frac{1}{2} \cdot \left(\frac{1}{2}-1\right) \right)\frac{1}{n^2} +o(\frac{1}{n^2}))}}=\lim_{n\to\infty}{n \ln{\left(1-\frac{1}{4n}\right)}}=-\frac{1}{4} $
so that  $\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}=\lim _{n \rightarrow \infty} e^{(1)}=e^{-\frac{1}{4}}$
this solution is right.
 A: We have
$$\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}=\lim_{n\to\infty} \frac1{n^n}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{1+k/n^2}}}\right)^{n}}$$
and
$$\frac{1}{\sqrt{1+k/n^2}}=1-\frac12\frac{k}{n^2}+O\left(\frac{k^2}{n^4}\right)$$
therefore
$$\sum_{k=1}^{n}{\frac{1}{\sqrt{1+k/n^2}}}=n-\frac12\frac{n(n+1)}{2n^2}+O\left(\frac1n\right)$$
and
$$\frac1{n^n}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{1+k/n^2}}}\right)^{n}}=\left(1-\frac{n+1}{4n^2}+O\left(\frac1{n^2}\right)\right)^n\to\frac1{\sqrt[4] e}$$
A: Another solution: Let $$S_n= \sum_{k=1}^{n} \frac{1}{n}\frac{1}{\sqrt{1+\frac{k}{n^2}}}$$
 We will prove that $$ \frac{2x+2}{2+x}<\sqrt{1+x} < 1+\frac{x}{2}, x>0~~~~(1)$$
 Right one: is nothing but $$2\sqrt{1+x}=-1-(1+x)-[1-\sqrt{1+x}]^2  \le 0.$$
 The left one isnothing but $$\frac{2(1+x)}{1+1+x} < \sqrt{1+x} \Rightarrow 2\sqrt{1+x}<1+(1+x)  \Rightarrow -[1-\sqrt{1+x}]<0.$$
 From  (1) it follows that $$ 1+ \frac{k}{2n^2+k} ~<~\sqrt{1+\frac{k}{n^2}}~ < ~1+\frac{k}{2n^2}$$
$$\Rightarrow 1+ \frac{k}{2n^2+\underline{2n}} ~<~\sqrt{1+\frac{k}{n^2}}~ < ~1+\frac{k}{2n^2}$$
$$\Rightarrow \left(1+ \frac{k}{2n^2+2n}\right)^{-1} ~>~\frac{1}{\sqrt{1+\frac{k}{n^2}}}~ > ~\left( 1+\frac{k}{2n^2} \right)^{-1}$$
$$\Rightarrow \left(1- \frac{k}{2n^2+2n}\right) ~>~\frac{1}{\sqrt{1+\frac{k}{n^2}}}~ > ~\left( 1-\frac{k}{2n^2} \right)$$
$$\Rightarrow  1- \sum_{k=1}^{n} \frac{k}{n(2n^2+2n)} ~>~\sum_{k=1}^{n} \frac{1}{n}\frac{1}{\sqrt{1+\frac{k}{n^2}}}~ >~ 1-\sum_{k=1}^{n}\frac{k}{2n^3}$$
$$\Rightarrow 1-\frac{1}{4n}~ > ~S_n~ >~1-\frac{1}{4n}-\frac{1}{4n^2}.$$
 Now $$\ln L= n \lim_{n \rightarrow \infty} \ln S_n = \lim_{n \rightarrow \infty} n \ln \left(1-\frac{1}{4n}\right) =\lim_{n \rightarrow \infty} n \frac{-1}{4n}=\frac{-1}{4} \Rightarrow L = e^{-\frac{1}{4}}$$ 
A: We can use the sandwich theorem twice for solving the limit
$$\text { Observe that } 
    \frac{n}{\sqrt{n^2 +n}}\le\alpha_n=\sum_{k=1}^{n}\frac{1}{\sqrt{n^2 +k}}\le\frac{n}{\sqrt{n^2+1}}, \text{ So, } \lim_{n\to\infty}\alpha_n=1.\space 
 \text { Proceeding as an usual case of } 1^{\infty}, $$
$$ \alpha=\lim_{n\to\infty}\left(\sum_{k=1}^{n}\frac{1}{\sqrt{n^2 +k}}\right) ^n=e^{\lim_{n\to\infty} n(\alpha_n -1)}=e^{\lim_{n\to\infty}\sum_{k=1}^{n}\left(\frac{n}{\sqrt{n^2+k}}  -1\right)}=e^{\lim_{n\to\infty}\sum_{k=1}^{n}\left(\frac{-k}{(n+\sqrt{n^2+k})\sqrt{n^2+k}}  \right)}$$
$$\text{ Now, use Sandwich theorem again and obtain that } \alpha=\frac{1}{\sqrt[4]{e}}$$
A: Following your solution from here
$$...= \lim_{n\to\infty}{n\ln{n \cdot2 \left(\sqrt{1+\frac{1}{n}}-1\right)}}=...$$
we have that
$$\sqrt{1+\frac{1}{n}}-1=\frac1{2n}-\frac1{8n^2}+O\left(\frac1{n^3}\right)$$
and therefore
$$\lim_{n\to\infty}{n\ln{n \cdot2 \left(\sqrt{1+\frac{1}{n}}-1\right)}}=\lim_{n\to\infty}{n\ln{\left(1-\frac1{4n}+O\left(\frac1{n^2}\right)\right)}}=-\frac14$$
