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Math Help - Limit of factorial in Alternating Series

  1. #1
    DBA
    DBA is offline
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    Limit of factorial in Alternating Series

    I need to find out if the Alternating Series is convergent or divergent by the Alternating Series Test.

    One condition is to check if the limit is zero.
    I have problems finding the limit, since I have a term (2k-1)! in the denominator and I do not know how to handle this...


    <br />
\Sigma^{\infty}_{k = 1} (-1)^{k-1}\frac{k!}{(2k-1)!}<br />

    <br />
\lim_{k \to \infty}\frac{k!}{(2k-1)!}<br />

    How can I write the fraction, so that things cancel out?

    <br />
\frac{k!}{(2k-1)!} = ?<br />

    For the following example I was able to write one term:

    <br />
\frac{k!}{(k+1)!} = \frac{k (k-1)(k-2)(k-3)(k-4) ... (2)(1)}{(k+1)(k)(k-1)(k-2)(k-3)(k-4)....(2)(1)} = \frac{1}{k+1}<br />

    Can someone please let me know how to do this with

    <br />
\frac{k!}{(2k-1)!}<br />

    Thanks!
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  2. #2
    MHF Contributor

    Joined
    Aug 2008
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    Paris, France
    Posts
    1,174
    Quote Originally Posted by DBA View Post
    I need to find out if the Alternating Series is convergent or divergent by the Alternating Series Test.

    One condition is to check if the limit is zero.
    I have problems finding the limit, since I have a term (2k-1)! in the denominator and I do not know how to handle this...


    <br />
\Sigma^{\infty}_{k = 1} (-1)^{k-1}\frac{k!}{(2k-1)!}<br />

    <br />
\lim_{k \to \infty}\frac{k!}{(2k-1)!}<br />

    How can I write the fraction, so that things cancel out?

    <br />
\frac{k!}{(2k-1)!} = ?<br />

    For the following example I was able to write one term:

    <br />
\frac{k!}{(k+1)!} = \frac{k (k-1)(k-2)(k-3)(k-4) ... (2)(1)}{(k+1)(k)(k-1)(k-2)(k-3)(k-4)....(2)(1)} = \frac{1}{k+1}<br />

    Can someone please let me know how to do this with

    <br />
\frac{k!}{(2k-1)!}<br />

    Thanks!
    You can do the same here and get \frac{k!}{(2k-1)!}=\frac{1}{(k+1)(k+2)\cdots(2k-1)} as soon as k\leq 2k-1, i.e. k\geq 1.
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