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Math Help - [SOLVED] Convergence....like all the others

  1. #1
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    [SOLVED] Convergence....like all the others

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

    Unlike some other that can do this one in there heads, I have to workout it out visually, I used the ration test, I believe I may have made some mistakes in simplification so I am asking for a check like always.


    a_n=\frac{(n!)^2}{(2n!)}

    a_n+1=\frac{(n+1!)^2}{(2n+2!)}

    \lim_{n\to\infty}\|\frac{(n+1!)^2}{(2n+2!)}\cdot\f  rac{(2n!)}{(n!)^2}

    Doing some simplification to

    \lim_{n\to\infty}\frac{(n+1)^2}{(2n+2)(2n+1)}

    Multiplying out the terms in the numerator and denominator gave me a fraction with the same degree in both denominator and numerator


    \lim_{n\to\infty}\frac{(n^2+2n+1)}{(4n^2+6n+2)}

    I can take the ratio of the degrees of the highest powers since they the same to find limit which gives L = \frac{1}{4} and the ratio test states, if the limit is less than 1, the series converges, is this correct?

    I just got confused when wolfram, simplified the series to \sum_{n=1}^{\infty}\frac{n!}{2} and stated it diverges, I was using wolfram for a quick check, If I did things correct or not.
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  2. #2
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    This looks right, if your series is  \sum_{n = 1}^{\infty} \frac{(n!)^2}{(2n)!} . That is, 2n factorial on the bottom as opposed to 2 times n factorial. Include the parenthesis around the 2n and you should get that the series converges to about 0.7364.

    Otherwise, wolfram is right and your series diverges for sure.
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  3. #3
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    Yes, the series was with 2n! not 2 * n!, thanks for the confirmation
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