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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0Thanks! I see now. You might post a reply to your question. – 2012-08-11
1 Answers
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This is a standard problem, this integral is called Integral of a Gaussian function
Let
$I:=\int_0^\infty e^{-x^2/2}dx = \frac{1}{2} \int_{\infty}^\infty e^{-x^2/2}dx$
Then
$(2I)^2= \int_{\infty}^\infty e^{-x^2/2}dx \int_{\infty}^\infty e^{-y^2/2}dy= \int \int _{R^2} e^{-x^2/2} e^{-y^2/2}dA \,.$
Use polar coordinates now.
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0"Am I need to solve the last integration? Then how can I solve it?" – 2012-08-03