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 $e^{1/3}$.

Using the fact that $\log(1+x)=x-\frac{x^2}{2}+\mathcal{O}(x^3)$ for small $x$, we find
\begin{align*}
\left( {\prod\limits_{r = 1}^n {\left( {1 + \frac{r}{{n^2 }}} \right)} } \right)^n  & = \exp \left( {n\sum\limits_{r = 1}^n {\log \left( {1 + \frac{r}{{n^2 }}} \right)} } \right) \\ &= \exp \left( {n\left( {\sum\limits_{r = 1}^n {\left( {\frac{r}{{n^2 }} - \frac{1}{2}\frac{{r^2 }}{{n^4 }} + \mathcal{O}\!\left( {\frac{{r^3 }}{{n^6 }}} \right)} \right)} } \right)} \right)
\\ &
 = \exp \left( {\frac{1}{n}\sum\limits_{r = 1}^n r  - \frac{1}{{2n^3 }}\sum\limits_{r = 1}^n {r^2 }  + \mathcal{O}\!\left( {\frac{1}{{n^5 }}} \right)\sum\limits_{r = 1}^n {r^3 } } \right) \\ &= \exp \left( {\frac{{n + 1}}{2} - \frac{{(n + 1)(2n + 1)}}{{12n^2 }} + \mathcal{O}\!\left( {\frac{1}{n}} \right)} \right)
\\ &
 = \exp \left( {\frac{n}{2} + \frac{1}{3} + \mathcal{O}\!\left( {\frac{1}{n}} \right)} \right).
\end{align*}
The result now follows.
