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?
Find $ \lim_{k \to \infty} \lim_{n \to \infty}\frac{1}{n}\sum_{m = 1}^{nk}\exp(-\frac{1}{2}\frac{m^2}{n^2})$
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calculus
integration
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0For the evaluation of the integral see [this question](http://math.stackexchange.com/q/9286/752). – 2012-08-03
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0Is my procedure is correct? can this problem solve in any other way? – 2012-08-03
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1I think your procedure is correct. Evaluating the integral is another matter and there are several ways to achieve this. – 2012-08-03
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0It 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 $$\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{m=1}^{nk}e^{-\frac{m^{2}}{2n^{2}}}=\lim_{n\rightarrow \infty }\frac{k}{n}\sum_{m=1}^{n}e^{-\frac{m^{2}k^{2}}{2n^{2}}}\ ?$$ – 2012-08-10
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0Put $nk=p$ 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. – 2012-08-11
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0Thanks! I see now. You might post a reply to your question. – 2012-08-11