Is there any proof for this formula $\lim_{n \to \infty} \prod_{k=1}^n \left (1+\frac {kx}{n^2} \right) =e^{x/2}$? Some times ago, In a mathematical problem book I sow that this formula. I don't no whether it is true or not. But now I'm try to prove it. I have no idea how to begin it. Any hint or reference would be appreciate. Thank you.
$$\lim_{n \to \infty} \prod_{k=1}^n \left (1+\frac {kx}{n^2} \right) =e^{x/2}$$

Here I use the method suggested by Daniel Fischer in the comment section:
Take the logarithm of  $$\prod _{k=1}^n \left(1+\dfrac{kx}{n^2} \right).$$ Then we have
$$\ln\left(\prod_{k=1}^n\ \left(1+\dfrac{kx}{n^2 }\right) \right)=\sum_{k=1}^n\ln\left(1+\dfrac{kx}{n^2 }\right).$$
By the Taylor series expansion of logarithms,
$$ \ln\left(1+\dfrac{kx}{n^2} \right)=\sum_{m=1}^\infty\dfrac{(-1)^{m-1}}{m}\left(\dfrac{ kx}{n^2 }\right)^m \\=
\left(\dfrac{kx}{n^2 }\right)-\dfrac{1}{2} \left(\dfrac{kx}{n^2 }\right)^2+\dfrac{1}{3} \left(\dfrac{kx}{n^2 }\right)^3-\dfrac{1}{4} \left(\dfrac{kx}{n^2 }\right)^4+\dfrac{1}{5} \left(\dfrac{kx}{n^2 }\right)^5-\dfrac{1}{6} \left(\dfrac{kx}{n^2 }\right)^6+ …$$
Now
$$\sum_{k=1}^n\ln\left(1+\dfrac{kx}{n^2} \right) =\dfrac{x}{n^2 }\left( \sum_{k=1}^nk\right)-\dfrac{x^2}{2n^4 }\left( \sum_{k=1}^nk^2\right)+\dfrac{x^3}{3n^6 } \left(\sum_{k=1}^nk^3 \right)-\dfrac{x^4}{4n^8 }\left(\sum_{k=1}^nk^4 \right)+ …$$
Note that
$$\sum_{k=1}^nk^r =\dfrac{n^{r+1}}{r+1}+\dfrac{n^r}{2} + O(n^{r-1})$$
is a $(r+1)^{\text{th}}$ degree polynomial of $n.$
Therefore
$$\lim_{n\to\infty}\left(\dfrac{1}{n^{2r}}  \sum_{k=1}^nk^r\right)=\left\{
\begin{array}{rl}
\frac{1}{2}:&r=1\\
0:&r>1
\end{array}
\right.
$$
Hence we can obtain $$\lim_{n\to\infty}\ln\left(\prod_{k=1}^n\left(1+\dfrac{kx}{n^2 }\right) \right)=\lim_{n\to\infty}\sum_{k=1}^n\ln\left(1+\dfrac{kx}{n^2 }\right)=\dfrac{x}{2}-0+0-0+ ...$$
$$\ln \left(\lim_{n\to\infty}\prod_{k=1}^n\left(1+\dfrac{kx}{n^2 }\right)\right)=\dfrac{x}{2}$$
$$\lim_{n\to\infty}\prod_{k=1}^n\left(1+\dfrac{kx}{n^2 }\right)=e^{x/2}$$
