# Thread: What is the limit? (infinite sum)

1. ## 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?

2. ## Re: What is the limit? (infinite sum)

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

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

3. ## Re: What is the limit? (infinite sum)

Granted, this may be nitpicking:

4. ## Re: What is the limit? (infinite sum)

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.

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