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Math Help - [SOLVED] Convergence Series

  1. #1
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    [SOLVED] Convergence Series

    The set of real numbers x for which the series \sum_{n=1}^{\infty}\frac{n!x^{2n}}{n^n(1+x^{2n})} converges is?

    How should this be approached?
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  2. #2
    Junior Member allsmiles's Avatar
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    well do know how to solve these types of problems or are they new for you?
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    Take note: \left( {\sqrt[n]{{\frac{{n!}}<br />
{{n^n }}}}} \right) \to \frac{1}<br />
{e}
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  4. #4
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    Quote Originally Posted by Plato View Post
    Take note: \left( {\sqrt[n]{{\frac{{n!}}<br />
{{n^n }}}}} \right) \to \frac{1}<br />
{e}
    So use the root test and then test values of x\in\mathbb{R}?
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  5. #5
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    Quote Originally Posted by allsmiles View Post
    well do know how to solve these types of problems or are they new for you?
    I know how to solve series problems. The issue here is method identifying the interval for x\in\mathbb{R}
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  6. #6
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    \lim_{n\to\infty}\sqrt[n]{\frac{n!x^{2n}}{n^n(1+x^{2n})}}=\sqrt[n]{\frac{n!}{n^n}}\frac{(x^{2n})^{\frac{1}{n}}}{(1+x  ^{2n})^{\frac{1}{n}}}=\frac{x^2}{e(1+x^{2n})^{\fra  c{1}{n}}}

    (1+x^{2n})^{\frac{1}{n}} this term is posing some issues now.

    To use L'Hopitals Rule, I need either \frac{\infty}{\infty},\ \frac{0}{0}, \ 0^0,\ \infty^{\infty}, but I have \infty^0.
    Last edited by dwsmith; May 20th 2010 at 04:21 PM.
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    Correction: You need either  \frac{\infty}{\infty}, \frac{0}{0}, 0^0, 1^\infty or \infty^0

     \infty^\infty is not an indeterminate.
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    \lim_{n\to\infty}\frac{x^2}{e}\left[(1+x^{2n})^{\frac{-1}{n}}\right]=\frac{x^2}{e}\lim_{n\to\infty}(1+x^{2n})^{\frac{-1}{n}}

    Then just exponentiate to convert to \frac{\infty}{\infty}? Or can I just ln (1+x^{2n})^{\frac{-1}{n}} WLOG by leaving \frac{x^2}{e} on the outside?
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    Yea I'd do natural log because  \lim_{n\to\infty}\frac{x^2}{e}\left[(1+x^{2n})^{\frac{-1}{n}}\right] = \lim_{n\to\infty} \frac{x^2}{e} * \lim_{n\to\infty}(1+x^{2n})^{\frac{-1}{n}}
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  10. #10
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    Quote Originally Posted by lilaziz1 View Post
    Yea I'd do natural log because  \lim_{n\to\infty}\frac{x^2}{e}\left[(1+x^{2n})^{\frac{-1}{n}}\right] = \lim_{n\to\infty} \frac{x^2}{e} * \lim_{n\to\infty}(1+x^{2n})^{\frac{-1}{n}}
    But can I ln only the right portion and let the left portion be?
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  11. #11
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    Yea. Remember to e the answer you get from  \lim_{n\to\infty}(1+x^{2n})^{\frac{-1}{n}} and multiply your final answer with  \frac{x^2}{e}
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  12. #12
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by dwsmith View Post
    But can I ln only the right portion and let the left portion be?
    Yes...

     \frac{x^2}{e} * \lim_{n\to\infty} (1+x^{2n})^{\frac{-1}{n}}

    Let  y = (1+x^{2n})^{\frac{-1}{n}}

     lny = \frac{-1}{n} ln ( 1+x^{2n} )

     \lim_{ n \to \infty } lny = \lim_{ n \to \infty } \frac{-1}{n} ln ( 1+x^{2n} ) = \lim_{ n \to \infty } \frac{-1}{n} \lim_{ n \to \infty } ln ( 1+x^{2n} ) = 0 * \infty

    The above is not defined but we note that the first function goes towards 0 faster then the second function goes toward infinity. So I would say this would approach 0.

    With this result we can use...

     \frac{x^2}{e} * \lim_{n\to\infty} (1+x^{2n})^{\frac{-1}{n}} =\frac{x^2}{e} * \lim_{n\to\infty} y = \frac{x^2}{e} * \lim_{n\to\infty} e^{lny} = x^2

    I think the above is valid...but i'm about 7 beers deep by now...soo dont trust me to much
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    Quote Originally Posted by AllanCuz View Post
    Yes...

     \frac{x^2}{e} * \lim_{n\to\infty} (1+x^{2n})^{\frac{-1}{n}}

    Let  y = (1+x^{2n})^{\frac{-1}{n}}

     lny = \frac{-1}{n} ln ( 1+x^{2n} )

     \lim_{ n \to \infty } lny = \lim_{ n \to \infty } \frac{-1}{n} ln ( 1+x^{2n} ) = \lim_{ n \to \infty } \frac{-1}{n} \lim_{ n \to \infty } ln ( 1+x^{2n} ) = 0 * \infty

    The above is not defined but we note that the first function goes towards 0 faster then the second function goes toward infinity. So I would say this would approach 0.

    With this result we can use...

     \frac{x^2}{e} * \lim_{n\to\infty} (1+x^{2n})^{\frac{-1}{n}} =\frac{x^2}{e} * \lim_{n\to\infty} y = \frac{x^2}{e} * \lim_{n\to\infty} e^{lny} = x^2

    I think the above is valid...but i'm about 7 beers deep by now...soo dont trust me to much
    That must be why you answered the question then because I don't need someone to solve it. My questions were on rules of certain operations.
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  14. #14
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by dwsmith View Post
    That must be why you answered the question then because I don't need someone to solve it. My questions were on rules of certain operations.
    oh why would you need to know that? Clearly if our limit is of some variable to some number, then all multiplication that has nothing to do with that variable can be taken outside of the limit.

    It's no different then....

     \lim_{x \to 0 } cf(x) = c \lim_{x \to 0 } f(x)

    Which you most definately know, so I'm confused at why you are confused?
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  15. #15
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    Quote Originally Posted by dwsmith View Post
    Can I just ln (1+x^{2n})^{\frac{-1}{n}} WLOG by leaving \frac{x^2}{e} on the outside?
    This was the question.
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