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Infinite series problem

  1. Jun 14, 2013 #1
  2. jcsd
  3. Jun 14, 2013 #2


    Staff: Mentor

  4. Jun 14, 2013 #3
    I just rewrote the thing in the link to only include the part I was having difficulty with.

    $$ \sum_{n = 0}^{\infty}\frac{1}{2^{n+1}}\frac{n}{n+1} = \sum_{n = 0}^{\infty}\frac{1}{2^{n+1}}\left(1 - \frac{1}{n+1}\right)=1-\sum_{n=0}^{\infty}\frac{1}{(n+1)2^{n+1}}=1-\sum_{n=1}^{\infty}\frac{1}{n2^{n}}$$. The link says that this is equal to $$1-\textstyle\log 2$$ (or less ambiguously, ln 2) which is why I said what I did in the first post. And I'm pretty sure it is true. I wrote a program and I summed from 1 to 100 and I got almost exactly ln 2.
  5. Jun 14, 2013 #4
  6. Jun 14, 2013 #5


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    Consider the geometric series:

    [tex]\sum_{n=0}^{+\infty} x^n = \frac{1}{1-x}[/tex]

    Then integrating yields

    [tex]\int_0^t \sum_{n=0}^{+\infty} x^n dx = \sum_{n=0}^{+\infty} \int_0^t x^n dx = \sum_{n=0}^{+\infty} \frac{1}{n+1} t^{n+1} = \sum_{n=1}^{+\infty} \frac{t^n}{n}[/tex]

    And thus

    [tex]\sum_{n=1}^{+\infty} \frac{t^n}{n} = \int_0^t\frac{1}{1-x} dx = -\log|1-t|[/tex]

    Filling in ##t=1/2## gives us

    [tex]\sum_{n=1}^{+\infty} \frac{1}{n2^n} = - \log(1/2) = \log(2)[/tex]

    I leave it up to you to justify each step in this calculation rigorously.
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