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Math Help - Summation of (-1)^n/(n+(-1)^n+1)

  1. #1
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    Summation of (-1)^n/(n+(-1)^n+1)

    Does the series in the attached image converge?

    This is my attempt at a solution:

    We have:
    -1/2 + 1 -1/4 + 1/3 -1/6 + 1/5 -.....

    I had this in an exam paper and answered that the series converges because the series:
    1 - 1/2 + 1/3 - 1/4 + ...
    converges (according to Liebniz).

    But I only got 2/10 for the question because rearranging an infinite series can change the result. So how do I prove this series does or does not converge?

    Thanks
    Attached Thumbnails Attached Thumbnails Summation of (-1)^n/(n+(-1)^n+1)-5b.jpg  
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  2. #2
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    Opalg's Avatar
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    Re: Summation of (-1)^n/(n+(-1)^n+1)

    Quote Originally Posted by durrrrrrrr View Post
    Does the series in the attached image converge?

    This is my attempt at a solution:

    We have:
    -1/2 + 1 -1/4 + 1/3 -1/6 + 1/5 -.....

    I had this in an exam paper and answered that the series converges because the series:
    1 - 1/2 + 1/3 - 1/4 + ...
    converges (according to Liebniz).

    But I only got 2/10 for the question because rearranging an infinite series can change the result. So how do I prove this series does or does not converge?
    Let S_n denote the sum of the first n terms of the series. Then S_{2n} is the same as the sum of the first 2n terms of the Leibniz series (because there is no problem with rearranging the terms in a finite sum). So S_{2n} converges to the same limit L as the Leibniz series.

    Next, the (2n+1)th term in the series is -\tfrac1{2(n+1)}, which converges to 0 as n\to\infty. Therefore S_{2n+1} = S_{2n} -\tfrac1{2(n+1)} \to L+0 = L.

    Finally, since S_{2n} and S_{2n+1} converge to the same limit L, it follows that the whole series converges to L.
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  3. #3
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    Re: Summation of (-1)^n/(n+(-1)^n+1)

    Quote Originally Posted by Opalg View Post
    Finally, since S_{2n} and S_{2n+1} converge to the same limit L, it follows that the whole series converges to L.
    Thanks, is this last stage true by induction?
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