What is the limit? (infinite sum)

Hello.

I am wondering about this limit

$\displaystyle \mathop {\lim }\limits_{x \to \infty } \sum\limits_{n = - \infty }^{ + \infty } {{{\left( { - 1} \right)}^n}{e^{ - 2{n^2}{x^2}}}} $

The answer is definitely $\displaystyle 1$ since this is related to Kolmogorov distribution which is recognized in statistics field.

But how to prove it?

Re: What is the limit? (infinite sum)

Oh silly me. Since the sum and all limits are convergent one can exchange those two processes...

$\displaystyle \mathop {\lim }\limits_{x \to \infty } \sum\limits_{n = - \infty }^{ + \infty } {{{\left( { - 1} \right)}^n}{e^{ - 2{n^2}{x^2}}}} = \sum\limits_{n = - \infty }^{ + \infty } {\mathop {\lim }\limits_{x \to \infty } {{\left( { - 1} \right)}^n}{e^{ - 2{n^2}{x^2}}}} = \mathop {\lim }\limits_{x \to \infty }{e^{ - 2 \cdot {0^2}{x^2}}}=1$

1 Attachment(s)

Re: What is the limit? (infinite sum)

Granted, this may be nitpicking:

Attachment 26547

Re: What is the limit? (infinite sum)

Quote:

Originally Posted by

**Pranas** Oh silly me. Since the sum and all limits are convergent one can exchange those two processes...

No, that's not true. One can interchange the processes only if they are **uniformly** convergent.

Quote:

$\displaystyle \mathop {\lim }\limits_{x \to \infty } \sum\limits_{n = - \infty }^{ + \infty } {{{\left( { - 1} \right)}^n}{e^{ - 2{n^2}{x^2}}}} = \sum\limits_{n = - \infty }^{ + \infty } {\mathop {\lim }\limits_{x \to \infty } {{\left( { - 1} \right)}^n}{e^{ - 2{n^2}{x^2}}}} = \mathop {\lim }\limits_{x \to \infty }{e^{ - 2 \cdot {0^2}{x^2}}}=1$