Calculate the limit $\lim_{n\rightarrow\infty}\left(\frac{\prod_1^n(1 +r/n^2)}{e^{1/2}}\right)^n$ How to calculate the limit
$$\lim_{n\rightarrow\infty}\left(\frac{\prod\limits_{r=1}^n\left(1 +\frac{r}{n^2}\right)}{e^{1/2}}\right)^n?$$
I tried using the sandwich theorem by taking logarithm of the limit and using the expansion
$$ x - (x^2)/2<\ln(1+x)<x$$
but I was getting different from answer from the two sides. The correct answer is $e^{1/3}$.

Hint:
If we take
$\begin{align}
&P(x) := \prod_{r=1}^n\left(x +\frac{r}{n^2}\right) \\[1mm]
&= x^n + \frac{\sum_{r_1=1}^n r_1}{n^2}x^{n-1} + \frac{\sum_{r_1=1}^n\sum_{r_2=1}^n r_1r_2}{n^4}x^{n-2} + \dots + \frac{\sum_{r_1=1}^n\sum_{r_2=1}^n\dots\sum_{r_n=1}^n r_1r_2\cdot...\cdot r_n}{n^{2n}} \\[1mm]
&= x^n + \frac{\sum_{r_1=1}^n r_1}{n^2}x^{n-1}  + \frac{\left(\sum_{r_1=1}^n r_1\right)^2}{n^4}x^{n-2} + \dots + \frac{\left(\sum_{r_1=1}^n r_1\right)^n}{n^{2n}} \\[1mm]
&= x^{n} + \frac{S}{n^2}x^{n-1} + \frac{S^2}{n^4}x^{n-2} + \dots + \frac{S^n}{n^{2n}} \tag{$\star$} \end{align}$
where $$S = \sum_{r_1=1}^n r_1 = \frac{n(n+1)}{2}$$
Then the numerator of the ratio of the limit
$$\lim_{n\rightarrow\infty}\left(\frac{\prod_{r=1}^n(1 +r/n^2)}{e^{1/2}}\right)^n$$
will be
$$\begin{align}
P(1) &= 1 + \frac{S}{n^2} + \frac{S^2}{n^4} + \dots + \frac{S^n}{n^{2n}} \\[1mm]
&= \frac{1 - \frac{S^{n+1}}{n^{2(n+1)}}}{1 - \frac{S}{n^2}} \\[1mm]
&= \frac{n\left(-n^{-n-1}\left(n+1\right)^{n+1}+2^{n+1}\right)}{2^n\left(n-1\right)}\\[1mm]
&= \frac{n}{n-1}\frac{\left(2^{n+1} - \left(1+\frac{1}{n}\right)^{n+1}\right)}{2^n} \\[1mm]
&\sim 2 - \frac{1}{2^n}\left(1+\frac{1}{n}\right)^{n}\end{align}$$
