What is the limit? (infinite sum)

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January 12th 2013, 04:31 AM
Pranas

What is the limit? (infinite sum)

Hello.

I am wondering about this limit
$\mathop {\lim }\limits_{x \to \infty } \sum\limits_{n = - \infty }^{ + \infty } {{{\left( { - 1} \right)}^n}{e^{ - 2{n^2}{x^2}}}}$

The answer is definitely $1$ since this is related to Kolmogorov distribution which is recognized in statistics field.
But how to prove it?
January 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...

$\mathop {\lim }\limits_{x \to \infty } \sum\limits_{n = - \infty }^{ + \infty } {{{\left( { - 1} \right)}^n}{e^{ - 2{n^2}{x^2}}}} = \sum\limits_{n = - \infty }^{ + \infty } {\mathop {\lim }\limits_{x \to \infty } {{\left( { - 1} \right)}^n}{e^{ - 2{n^2}{x^2}}}} = \mathop {\lim }\limits_{x \to \infty }{e^{ - 2 \cdot {0^2}{x^2}}}=1$
January 12th 2013, 05:34 PM
johng

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Re: What is the limit? (infinite sum)

Granted, this may be nitpicking:
Attachment 26547
January 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:

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