Find the limit of $(a_k)_k$ where $a_k = \lim_{n\to\infty} \frac1n\sum_{m=1}^{kn} \mathrm{exp}(\frac12.\frac{m^2}{n^2})$

Question

For $k \ge 1$, let $$a_k = \lim_{n\to\infty} \frac1n\sum_{m=1}^{kn} e^{\frac12.\frac{m^2}{n^2}}$$ Find the value of $$\lim_{k\to\infty} a_k$$

Answer 1

$$a_k=\lim_{n\to\infty}\frac1n\sum_{m=1}^{kn}e^{^{-\tfrac12\cdot\tfrac{m^2}{n^2}}}=\int_0^ke^{^{-\frac{x^2}2}}dx\qquad=>\qquad\lim_{k\to\infty}a_k=\int_0^\infty e^{^{-\frac{x^2}2}}dx=\sqrt{\frac\pi2}$$