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For $ k\geq 1$, let $$a_k=\lim_{n \to \infty}\frac{1}{n}\sum_{m = 1}^{nk}\exp(-\frac{1}{2}\frac{m^2}{n^2})$$Find $$\lim_{k \to \infty}a_k.$$I proceed in this way: $$a_k=\lim_{n \to \infty}\frac{1}{n}\sum_{m = 1}^{nk}\exp(-\frac{1}{2}\frac{m^2}{n^2})=\int_0^ke^{-x^2/2}dx$$ So $$\lim_{k \to \infty}a_k=\int_0^\infty e^{-x^2/2}dx$$ Is this procedure is right . Am I need to solve the last integration? Then how can I solve it?

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For the evaluation of the integral see this question. – Américo Tavares Aug 3 '12 at 14:31
Is my procedure is correct? can this problem solve in any other way? – Argha Aug 3 '12 at 15:03
I think your procedure is correct. Evaluating the integral is another matter and there are several ways to achieve this. – DonAntonio Aug 3 '12 at 16:35
It seems to me that $$\int_{0}^{k}e^{-\frac{x^{2}}{2}}dx=\lim_{n\rightarrow \infty }\frac{k}{n}\sum_{m=1}^{n}e^{-\frac{m^{2}k^{2}}{2n^{2}}}.$$ But how do you show that $$\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{m=1}^{nk}e^{-\frac{m^{2}}{2n^{2}}}=\lim_{n\rightarrow \infty }\frac{k}{n}\sum_{m=1}^{n}e^{-\frac{m^{2}k^{2}}{2n^{2}}}\ ?$$ – Américo Tavares Aug 10 '12 at 16:59
Put $nk=p$ we get $$\frac{1}{n}\sum_{m=1}^{nk} e^{-\frac{m^2}{2n^2}}$$ $$=\frac{k}{p}\sum_{m=1}^{p} e^{-\frac{m^2 k^2}{2p^2}}$$Also note that as $n\to \infty$ $nk=p \to \infty$ also. – Argha Aug 11 '12 at 5:30

This is a standard problem, this integral is called Integral of a Gaussian function


$$I:=\int_0^\infty e^{-x^2/2}dx = \frac{1}{2} \int_{\infty}^\infty e^{-x^2/2}dx$$


$$(2I)^2= \int_{\infty}^\infty e^{-x^2/2}dx \int_{\infty}^\infty e^{-y^2/2}dy= \int \int _{R^2} e^{-x^2/2} e^{-y^2/2}dA \,.$$

Use polar coordinates now.

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I don't think this answers the OP's actual question – DonAntonio Aug 3 '12 at 16:34
"Am I need to solve the last integration? Then how can I solve it?" – N. S. Aug 3 '12 at 17:35

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