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

  1. #1
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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 $\displaystyle 1*3*5***$ means, and I do not know how to approach or solve the problem.
    $\displaystyle 1*3*5***(2n-1)$ means multiplication of first $\displaystyle n$ odd numbers from $\displaystyle 1$ to $\displaystyle (2n-1)$.
    In general it is shown as
    $\displaystyle \prod\limits_{k=1}^{n}(2k-1)$.
    I guess that your series is
    $\displaystyle \sum\limits_{n=1}^{\infty}\dfrac{1\times3\times\cd ots\times(2n-1)}{n!}$,
    which can be also represented as
    $\displaystyle \sum\limits_{n=1}^{\infty}\dfrac{1}{n!}\prod\limit s_{k=1}^{n}(2k-1)$.
    Your general term for this series is
    $\displaystyle a_{n}:=\dfrac{1\times3\times\cdots\times(2n-1)}{n!}$.
    So that, we have
    $\displaystyle \dfrac{a_{n+1}}{a_{n}}=\dfrac{\dfrac{1\times3\time s\cdots\times\big(2(n+1)-1\big)}{(n+1)!}}{\dfrac{1\times3\times\cdots\times (2n-1)}{n!}}$
    .........$\displaystyle =\dfrac{\dfrac{1\times3\times\cdots\times(2n+1)}{( n+1)!}}{\dfrac{1\times3\times\cdots\times(2n-1)}{n!}}$
    .........$\displaystyle =\dfrac{\dfrac{1\times3\times\cdots\times(2n-1)\times(2n+1)}{n!(n+1)}}{\dfrac{1\times3\times\cd ots\times(2n-1)}{n!}}$
    .........$\displaystyle =\dfrac{2n+1}{n}$
    By using the ration test, we get
    $\displaystyle \lim\limits_{n\to\infty}\dfrac{a_{n+1}}{a_{n}}=\li m\limits_{n\to\infty}\dfrac{2n+1}{n}$
    ..................$\displaystyle =2>1$,
    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...

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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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.

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

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

    Tonio
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