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

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?


$a_{k}$ is simply the Riemann sum of the the integral $\int_{0}^{k}e^{-\frac{x^{2}}{2}}\,dx$ corresponding to the partition $\{0,\frac{1}{n}, \frac{2}{n}, \cdots, \frac{kn}{n}\}$. Thus $a_{k} = \int_{0}^{k}e^{-\frac{x^{2}}{2}}\,dx$. So $\lim_{k \rightarrow \infty}a_{k} = \int_{0}^{\infty} e^{-\frac{x^{2}}{2}}\,dx = \sqrt{\frac{\pi}{2}}$.

  • $\begingroup$ I see that my answer is correct. Is my method correct?' $\endgroup$ – user554970 May 3 '18 at 17:05
  • $\begingroup$ I don't see how you are using the Central limit theorem and I think in the CLT you usually have a $\sqrt{n}$ instead of an $n$ in the denominator. $\endgroup$ – harmonicuser May 3 '18 at 17:10

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