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Math Help - series summation problem

  1. #1
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    series summation problem

    \sum _{n=0}^{\infty } \left(\frac{1}{2 (1+n) (1+2 n)}\right)=\text{Log}[2]
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: series summation problem

    Quote Originally Posted by capea View Post
    \sum _{n=0}^{\infty } \left(\frac{1}{2 (1+n) (1+2 n)}\right)=\text{Log}[2]
    From the series expansion...

    \sum_{n=0}^{\infty} x^{2n}= \frac{1}{1-x^{2}} (1)

    ... we derive first...

    \sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1} = \int_{0}^{x} \frac{d \xi}{1-\xi^{2}} = \frac{1}{2}\ \ln \frac{1+x}{1-x} (2)

    ... and then...

    \sum_{n=0}^{\infty} \frac{x^{2n+2}}{(2n+1)\ (2n +2)}= \frac{1}{2}\ \int_{0}^{x} \ln \frac{1+\xi}{1-\xi}\ d \xi =

    = \frac{1}{2}\ \{(1+x)\ \ln (1+x) + (1-x)\ \ln (1-x) \} (3)

    Now, setting in (3) x=1 You obtain...

    \sum_{n=0}^{\infty} \frac{1}{(2n+1)\ (2n +2)}= \ln 2

    Kind regards

    \chi \sigma
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  3. #3
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    Re: series summation problem

    thanks
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: series summation problem

    Quote Originally Posted by capea View Post
    \sum _{n=0}^{\infty } \left(\frac{1}{2 (1+n) (1+2 n)}\right)=\text{Log}[2]

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

    \sum {\frac{1}{2n+1}-\frac{1}{2(n+1)}}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{(-1)^{k+1}}{k}+...=\sum {\frac{(-1)^{n+1}}{n}}=\ln2
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