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Math Help - Testing a series for convergence - is my solution correct?

  1. #1
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    Testing a series for convergence - is my solution correct?

    Hello everyone,

    I tested the following series for convergence, but I'm not sure if all the steps I took are strictly legal. Thus I would like to ask you to take a look at them.

    \sum_{n=1}^{+\infty}{\frac{\left(n!\right)^2}{n^n}  }

    I use the following fact to construct a series that is less than the one above for a comparison test.

    n!>\sqrt{2\pi n}\left(\frac{n}{e}\right)^n

    \sum_{n=1}^{+\infty} {\frac{\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\right)^2}{n^n}}

    Now I conduct a root test for this series to test it for convergence.

    \lim\limits_{n\rightarrow+\infty} {\frac{\sqrt[n]{2\pi n\left(\frac{n}{e}\right)^{2n}}}{\sqrt[n]{n^n}}}<br />
= \lim\limits_{n\rightarrow+\infty} {\frac{\sqrt[n]{2\pi n}\left(\frac{n}{e}\right)^2}{n}}<br />
= \lim\limits_{n\rightarrow+\infty} {\sqrt[n]{2\pi n}\frac{n}{e^2}}<br />
= \lim\limits_{n\rightarrow+\infty} {e^{\frac{1}{n}\log{\left(2\pi n\right)}}\frac{n}{e^2}}<br />

    At this point the exponent of the exponential function is the problem, which I solve by using l'H˘pital's rule. (I'm not exactly sure if I am allowed to do this here.)

    \lim\limits_{n\rightarrow+\infty}{\frac{\log{\left  (2\pi n\right)}}{n}} = <br />
\lim\limits_{n\rightarrow+\infty} {\frac{2\pi}{2\pi n}}=\lim\limits_{n\rightarrow+\infty} {\frac{1}{n}}

    Now I move this newly obtained exponent back into the original limit calculation.

    \lim\limits_{n\rightarrow+\infty} {e^{\frac{1}{n}}\frac{n}{e^2}}=+\infty

    Since the smaller series diverges, also the greater one has to diverge.

    I would appreciate it, if someone could tell me if I am right.
    Last edited by thomasdotnet; September 7th 2009 at 02:46 AM. Reason: replaced ( and ) with the proper Latex equivalents \left( and \right)
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  2. #2
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    The l'Hopital rule can be applied to obtain the limit when x\to\infty, hence if the limit exist the discrete limit when n\to\infty also exists and agree. So this step is correct.

    Also calculating the limit of the general term and observing that is infinty will give you the divergence of the series.
    Last edited by Enrique2; September 6th 2009 at 01:39 PM. Reason: A mistake in the argument
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  3. #3
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    \sqrt[n]{{\frac{{\left( {n!} \right)^2 }}{{n^n }}}} = \sqrt[n]{{\frac{{\left( {n!} \right)}}{{n^n }}}}\sqrt[n]{{n!}}

    \left( {\sqrt[n]{{\frac{{\left( {n!} \right)}}{{n^n }}}}} \right) \to \frac{1}{e}
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  4. #4
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    Thank you very much for your replies!
    I found out that l'H˘pital's rule is not to be used for sequences, except if the sequence can be transformed into a function f(x) such that f(n)=a_n, which is fortunately the case here.

    Quote Originally Posted by Plato View Post
    \sqrt[n]{{\frac{{\left( {n!} \right)^2 }}{{n^n }}}} = \sqrt[n]{{\frac{{\left( {n!} \right)}}{{n^n }}}}\sqrt[n]{{n!}}

    \left( {\sqrt[n]{{\frac{{\left( {n!} \right)}}{{n^n }}}}} \right) \to \frac{1}{e}
    Thanks for that hint Plato, but how I can I prove that \sqrt[n]{{n!}}\rightarrow+\infty?
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