What is the limit? (infinite sum)

**Pranas** · January 12th 2013, 04:31 AM

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?

**Pranas** · January 12th 2013, 10:07 AM

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

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

**johng** · January 12th 2013, 05:34 PM

Granted, this may be nitpicking:

What is the limit? (infinite sum)-alternatingseries.png

**HallsofIvy** · January 12th 2013, 06:22 PM

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.

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

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