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Math Help - What is the limit? (infinite sum)

  1. #1
    Member Pranas's Avatar
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    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?
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  2. #2
    Member Pranas's Avatar
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    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
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  3. #3
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    Re: What is the limit? (infinite sum)

    Granted, this may be nitpicking:
    What is the limit? (infinite sum)-alternatingseries.png
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  4. #4
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    Re: What is the limit? (infinite sum)

    Quote Originally Posted by Pranas View Post
    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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