sequences and series - $a_k = limlimits_{nto infty} sumlimits_{m=1}^{m=kn} frac{1}{n} e^{frac{-m^2}{2n^2}}$

$\displaystyle a_k = \lim_{n\to \infty} \sum_{m=1}^{m=kn} \frac{1}{n} e^{\frac{-m^2}{2n^2}}$

Find $\displaystyle \lim_{k \to \infty} a_k$

I tried using integral test and it resulted to nothing . ultimately I took the help of Central limit Theorem , even though it is completely wrong .

$\displaystyle a_k = \lim_{n\to \infty} \sum_{m=1}^{m=kn} \frac{1(\sqrt{2\pi)}}{n\sqrt{2\pi}} e^{\frac{-(m-0)^2}{2n^2}}$

$\displaystyle \Rightarrow a_k = \lim_{n\to \infty} \sqrt{2\pi} \,\int_{m=1}^{m=kn} \frac{1}{n\sqrt{2\pi}} e^{\frac{-(m-0)^2}{2n^2}} \quad\quad\quad,\text{where m is N}(0,n^2) $

$\displaystyle \Rightarrow a_k= \lim_{n\to \infty}\sqrt{2\pi} P(1\le m \le kn)=\lim_{n\to \infty}\sqrt{2\pi} P\left(\frac{1}{n}\le \frac{m}{n} \le \frac{kn}{n} \right) $

$\displaystyle \Rightarrow a_k= \sqrt{2\pi} \lim_{n\to \infty}\left(\Phi(k)\,-\,\Phi\left(\frac{1}{n}\right)\right)=\sqrt{2\pi}\left( \Phi(k) -\frac{1}{2}\right)$

$\displaystyle \therefore \lim_{k \to \infty} a_k =\sqrt{\frac{\pi}{2}}$

Any helpful insights?