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Math Help - Power series, term-by-term integration

  1. #1
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    Lightbulb Power series, term-by-term integration

    Evaluate

    ∑ 1 / [n(n+1) 5^n]
    n=1
    (hint: start with a geometric series, then use term-by-term integration)
    ===================================

    Attempt:

    x^n = 1/(1-x) for |x|<1
    n=0

    Take the indefinite integral of both sides...

    x^{n+1} /(n+1) + C= -ln(1-x)
    n=0
    Put x=0 => C=0

    x^{n+1} /(n+1) = -ln(1-x)
    n=0

    Divide both sides by x^2

    x^{n-1} /(n+1) = - ln(1-x) / x^2
    n=0


    x^{n-1} /(n+1) = - [ln(1-x) / x^2] - 1/x
    n=1

    How to continue?

    Thanks for helping!
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  2. #2
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    Quote Originally Posted by kingwinner View Post
    Evaluate

    ∑ 1 / [n(n+1) 5^n]
    n=1
    (hint: start with a geometric series, then use term-by-term integration)
    ===================================

    Attempt:

    x^n = 1/(1-x) for |x|<1
    n=0

    Take the indefinite integral of both sides...

    x^{n+1} /(n+1) + C= -ln(1-x)
    n=0
    Put x=0 => C=0

    x^{n+1} /(n+1) = -ln(1-x)
    n=0

    Divide both sides by x^2

    x^{n-1} /(n+1) = - ln(1-x) / x^2
    n=0


    x^{n-1} /(n+1) = - [ln(1-x) / x^2] - 1/x
    n=1

    How to continue?

    Thanks for helping!

    I'd begin with \sum^\infty_{n=1}x^{n-1}=\frac{1}{1-x},\,|x|<1\Longrightarrow \sum^\infty_{n=1}\frac{x^n}{n}=-\ln(1-x),\,|x|<1 \Longrightarrow \sum^\infty_{n=1}\frac{x^{n+1}}{n(n+1)}=(1-x)\ln(1-x)+x-1,\,|x|<1 , and now just substitute x=\frac{1}{5} ...

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    I'd begin with \sum^\infty_{n=1}x^{n-1}=\frac{1}{1-x},\,|x|<1\Longrightarrow \sum^\infty_{n=1}\frac{x^n}{n}=-\ln(1-x),\,|x|<1 \Longrightarrow \sum^\infty_{n=1}\frac{x^{n+1}}{n(n+1)}=(1-x)\ln(1-x)+x-1,\,|x|<1
    How did you get the -1 at the end? Are you sure it is -1, instead of 0?

    , and now just substitute x=\frac{1}{5} ...
    But if we just substitute x=1/5, we will get 1/5^{n+1} which is not what we want. We need 1/5^n

    Thanks!
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  4. #4
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    Quote Originally Posted by kingwinner View Post
    How did you get the -1 at the end? Are you sure it is -1, instead of 0?

    A mistake, indeed. Erase it.


    But if we just substitute x=1/5, we will get 1/5^{n+1} which is not what we want. We need 1/5^n


    This is not problem at all: \frac{1}{5^{n+1}}=\frac{1}{5}\cdot\frac{1}{5^n} ...

    Tonio


    Thanks!
    .
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