$\lim_{n\rightarrow \infty}\left[\frac{\left(1+\frac{1}{n^2}\right)\cdot \cdots\cdots \left(1+\frac{n}{n^2}\right)}{\sqrt{e}}\right]^n$ 
$$\lim_{n\rightarrow \infty}\Bigg[\frac{\bigg(1+\frac{1}{n^2}\bigg)\bigg(1+\frac{2}{n^2}\bigg)\cdots\cdots \bigg(1+\frac{n}{n^2}\bigg)}{\sqrt{e}}\Bigg]^n$$

Try: $$y=\lim_{n\rightarrow \infty}\Bigg[\frac{\bigg(1+\frac{1}{n^2}\bigg)\bigg(1+\frac{2}{n^2}\bigg)\cdots\cdots \bigg(1+\frac{n}{n^2}\bigg)}{\sqrt{e}}\Bigg]^n$$
$$\log_{e}(y) =n\sum^{n}_{r=1}\log_{e}\bigg(1+\frac{r}{n^2}\bigg)-\frac{n}{2}$$
could some help me to solve it , thanks
 A: We have $$\log{\left ( 1+\frac{r}{n^2} \right )}=\frac{r}{n^2}-\frac12\frac{r^2}{n^4}+O\left(\frac{r^3}{n^6}\right),$$ that means
$$n\sum^n_ {r=1}\log{\left ( 1+\frac{r}{n^2} \right )}=\frac{n(n+1)}2\frac1n-\frac12\frac{n(n+1)(2n+1)}6\frac1{n^3}+O(n^4)\frac1{n^5},$$ so we get
$$n\sum^n_ {r=1}\log{\left ( 1+\frac{r}{n^2} \right )}-\frac{n}2\to\frac12-\frac16=\frac13$$ as $n\to\infty$.
A: We have that
$$\begin{align}
\left[\frac{\left(1+\frac{1}{n^2}\right)\cdots \left(1+\frac{n}{n^2}\right)}{\sqrt{e}}\right]^n&=\frac{\left(\frac{(n^2+n)!}{(n^2)!}\right)^n}{n^{2n^2} e^{n/2}}
\sim\frac{\left(1+\frac{1}{n}\right)^{n^3+n^2+\frac{n}{2}}}{e^{n^2+n/2}}
\\
&\sim\frac{\exp\left((n^3+n^2+\frac{n}{2})\ln\left(1+\frac{1}{n}\right)\right)}{e^{n^2+n/2}}\\
&\sim\frac{\exp\left((n^2+n+\frac{1}{2})-\frac{n+1}{2}+\frac{1}{3}\right)}{e^{n^2+n/2}}\to e^{1/3}
\end{align}$$
where we used the Stirling approximation  for factorials and the expansion $$\ln(1+t)= t-\frac{t^2}{2}+\frac{t^3}{3}+o(t^3).$$
A: Answer seems to be $e^{1/2}. If you set the e term aside and the treat the rest as a 1^infinity form of limit you finally get e^1/2 without using any approximations
