# What is the limit? (infinite sum)

• Jan 12th 2013, 04:31 AM
Pranas
What is the limit? (infinite sum)
Hello.

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

The answer is definitely $\displaystyle 1$ since this is related to Kolmogorov distribution which is recognized in statistics field.
But how to prove it?
• Jan 12th 2013, 10:07 AM
Pranas
Re: What is the limit? (infinite sum)
Oh silly me. Since the sum and all limits are convergent one can exchange those two processes...

$\displaystyle \mathop {\lim }\limits_{x \to \infty } \sum\limits_{n = - \infty }^{ + \infty } {{{\left( { - 1} \right)}^n}{e^{ - 2{n^2}{x^2}}}} = \sum\limits_{n = - \infty }^{ + \infty } {\mathop {\lim }\limits_{x \to \infty } {{\left( { - 1} \right)}^n}{e^{ - 2{n^2}{x^2}}}} = \mathop {\lim }\limits_{x \to \infty }{e^{ - 2 \cdot {0^2}{x^2}}}=1$
• Jan 12th 2013, 05:34 PM
johng
Re: What is the limit? (infinite sum)
Granted, this may be nitpicking:
Attachment 26547
• Jan 12th 2013, 06:22 PM
HallsofIvy
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:

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