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Math Help - Prove convergence and find the sum

  1. #1
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    Prove convergence and find the sum

    \sum^\infty_{n=1}\frac{(-1)^{n+1}}{2n}

    I can figure out how to prove this series converges, but I cannot figure out how to find its sum.

    I know this is the answer:
    \sum^\infty_{n=1}\frac{(-1)^{n+1}}{2n} = \frac{log(2)}{2}

    But how?

    I got a hint to use this: tan^nu du = \frac{1}{n-1}tan^{n-1}u-\int{tan^{n-2}}du

    ________________________________________________
    Proof it converges:

    b_n =\frac{1}{2n}
    b_{n+1}=\frac{-1}{2n^2}

    Since n=1, n is always positive,

    b_n > b_{n+1}

    ________________________________________________

    A hint to get me started would be awesome.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Cursed View Post
    \sum^\infty_{n=1}\frac{(-1)^{n+1}}{2n}

    I can figure out how to prove this series converges, but I cannot figure out how to find its sum.

    I know this is the answer:
    \sum^\infty_{n=1}\frac{(-1)^{n+1}}{2n} = \frac{log(2)}{2}

    But how?

    I got a hint to use this: tan^nu du = \frac{1}{n-1}tan^{n-1}u-\int{tan^{n-2}}du

    ________________________________________________
    Proof it converges:

    b_n =\frac{1}{2n}
    b_{n+1}=\frac{-1}{2n^2}

    Since n=1, n is always positive,

    b_n > b_{n+1}

    ________________________________________________

    A hint to get me started would be awesome.
    That hint is absolutely stupid.

    \frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^n\implies \int\frac{dx}{1+x}=\ln(1+x)=-\sum_{n=1}^{\infty}\frac{(-1)^nx^n}{n} ignoring all of the justification since I don't think you care about that.
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  3. #3
    Junior Member
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    Thanks.

    But how would I go about using that so-called "hint"? I figure the squeeze lemma and integral test would apply -- I just don't know how exactly I'd use them.
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