Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
1  
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

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.