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Math Help - Series Convergence question dealing with 1*3*5* and n!

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    Series Convergence question dealing with 1*3*5* and n!

    Determine whether series converges or diverges.

    (1*3*5***(2n-1))/n!

    I believe you use the ratio test. It is the algebra in this that I do not know how to do. I don't know what 1*3*5*** means, and I do not know how to approach or solve the problem.
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  2. #2
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by ihatemath View Post
    Determine whether series converges or diverges.

    (1*3*5***(2n-1))/n!

    I believe you use the ratio test. It is the algebra in this that I do not know how to do. I don't know what 1*3*5*** means, and I do not know how to approach or solve the problem.
    1*3*5***(2n-1) means multiplication of first n odd numbers from 1 to (2n-1).
    In general it is shown as
    \prod\limits_{k=1}^{n}(2k-1).
    I guess that your series is
    \sum\limits_{n=1}^{\infty}\dfrac{1\times3\times\cd  ots\times(2n-1)}{n!},
    which can be also represented as
    \sum\limits_{n=1}^{\infty}\dfrac{1}{n!}\prod\limit  s_{k=1}^{n}(2k-1).
    Your general term for this series is
    a_{n}:=\dfrac{1\times3\times\cdots\times(2n-1)}{n!}.
    So that, we have
    \dfrac{a_{n+1}}{a_{n}}=\dfrac{\dfrac{1\times3\time  s\cdots\times\big(2(n+1)-1\big)}{(n+1)!}}{\dfrac{1\times3\times\cdots\times  (2n-1)}{n!}}
    ......... =\dfrac{\dfrac{1\times3\times\cdots\times(2n+1)}{(  n+1)!}}{\dfrac{1\times3\times\cdots\times(2n-1)}{n!}}
    ......... =\dfrac{\dfrac{1\times3\times\cdots\times(2n-1)\times(2n+1)}{n!(n+1)}}{\dfrac{1\times3\times\cd  ots\times(2n-1)}{n!}}
    ......... =\dfrac{2n+1}{n}
    By using the ration test, we get
    \lim\limits_{n\to\infty}\dfrac{a_{n+1}}{a_{n}}=\li  m\limits_{n\to\infty}\dfrac{2n+1}{n}
    .................. =2>1,
    which shows that your series is divergent.
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by ihatemath View Post
    Determine whether series converges or diverges.

    (1*3*5***(2n-1))/n!

    I believe you use the ratio test. It is the algebra in this that I do not know how to do. I don't know what 1*3*5*** means, and I do not know how to approach or solve the problem.
    The general term of the series is...

    \displaystyle a_{n} = \frac{1}{1}\ \frac{3}{2}\ \frac{5}{3}... \frac{2n-1}{n} (1)

    ... and because is \displaystyle \lim_{n \rightarrow \infty} a_{n} = \infty the series [strongly] diverges...

    Kind regards

    \chi \sigma
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    Quote Originally Posted by ihatemath View Post
    Determine whether series converges or diverges.

    (1*3*5***(2n-1))/n!

    I believe you use the ratio test. It is the algebra in this that I do not know how to do. I don't know what 1*3*5*** means, and I do not know how to approach or solve the problem.

    \frac{1\cdot 3\cdot\ldots\cdot (2n-1)}{n!}=\frac{(2n)!}{2^nn!}.

    Now use the quotient test to prove at once that the series diverges.

    Tonio
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