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Math Help - Simple series

  1. #1
    MHF Contributor chiph588@'s Avatar
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    Simple series

    I'm not sure if this is worthy of the challenge section, but here goes.

    Without using Taylor series/generating functions, show  \displaystyle \sum_{n=1}^\infty \frac{(-1)^n}n = -\log2 .
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  2. #2
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    The MacLaurin Series for \displaystyle \ln{(1+x)} = \sum_{n = 1}^{\infty}\frac{(-1)^{n+1}x^n}{n} for \displaystyle -1 < x \leq 1.


    So \displaystyle \ln{2} = \ln{(1 + 1)} = \sum_{n = 1}^{\infty}\frac{(-1)^{n+1}\cdot 1^n}{n} = \sum_{n = 1}^{\infty}\frac{(-1)^{n+1}}{n}.


    And finally \displaystyle -\ln{2} = \frac{\ln{2}}{-1} = \frac{\sum_{n = 1}^{\infty}\frac{(-1)^{n+1}}{n}}{-1} = \sum_{n = 1}^{\infty}\frac{(-1)^n}{n}.


    Edit: Oops, didn't read the question fully stating not to use Taylor series hahaha. Ah well, it stays because it's beautiful
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  3. #3
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    Here's another...

    \displaystyle \ln{2} = \int_1^2{\frac{1}{x}\,dx}.

    Substitute \displaystyle x = 1 + t so that \displaystyle dx = dt. When \displaystyle x= 1, t = 0 and when \displaystyle x = 2, t = 1 and the integral becomes

    \displaystyle \int_0^1{\frac{1}{1 + t}\,dt}

    \displaystyle = \int_0^1{\frac{1}{1 - (-t)}\,dt}

    \displaystyle = \int_0^1{\sum_{n = 0}^{\infty}(-t)^n\,dt} provided \displaystyle |-t| < 1 \implies |t| < 1

    \displaystyle = \sum_{n = 0}^{\infty}{\int_0^1{(-t)^n\,dt}}

    \displaystyle = \sum_{n = 0}^{\infty}{\lim_{\epsilon \to 1}\int_0^{\epsilon}{(-t)^n\,dt}}.

    The rest should follow.
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  4. #4
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    \displaystyle = \int_0^1{\frac{1}{1 - (-t)}\,dt}

    \displaystyle = \int_0^1{\sum_{n = 0}^{\infty}(-t)^n\,dt} provided \displaystyle |-t| < 1 \implies |t| < 1
    Isn't \displaystyle \frac{1}{1+t} = \sum_{n\ge 0}(-1)^nt^n a Taylor series?
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Hints

    (i) \displaystyle\sum_{k=1}^n{\dfrac{1}{k}}=\log n+C+o(1)\;\textrm{as}\;n\rightarrow{+\infty}

    (ii) The following sequence is convergent

    S_m=\displaystyle\sum_{i=1}^m{\dfrac{(-1)^k}{k}}

    (iii) Express

    S_{2n}=\ldots=\displaystyle\sum_{k=1}^n{\dfrac{1}{  k}}-\displaystyle\sum_{k=1}^{2n}{\dfrac{1}{k}}

    (iv) Apply (i) to prove that

    \displaystyle\lim_{n \to{+}\infty}{S_{2n}}=-\log 2

    which is exactly the sum of the series.


    Fernando Revilla
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  6. #6
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    Quote Originally Posted by TheCoffeeMachine View Post
    Isn't \displaystyle \frac{1}{1+t} = \sum_{n\ge 0}(-1)^nt^n a Taylor series?
    No, it's a Geometric Series :P
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  7. #7
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by FernandoRevilla View Post

    (ii) The following sequence is convergent

    S_m=\displaystyle\sum_{i=1}^m{\dfrac{(-1)^k}{k}}

    (iii) Express

    S_{2n}=\ldots=\displaystyle\sum_{k=1}^n{\dfrac{1}{  k}}-\displaystyle\sum_{k=1}^{2n}{\dfrac{1}{k}}
    You've got the right idea, but this step is not correct.

    Quote Originally Posted by Prove It View Post
    No, it's a Geometric Series :P
    It's also a Taylor series too!
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  8. #8
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by chiph588@ View Post
    You've got the right idea, but this step is not correct.
    It is correct.

    Hint : Express

    S_{2n}=\displaystyle\sum_{k=1}^n{\dfrac{1}{2k}}-\displaystyle\sum_{k=1}^n{\dfrac{1}{2k-1}}


    Fernando Revilla
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