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Sum of k x^k?

  1. Jan 24, 2010 #1
    [solved] Sum of k x^k?

    I happened upon a thread in a math forum, where someone asserted that this is true:

    [tex]\sum_{k=0}^\infty (k+1) \left(\frac{5}{6}\right)^k = 36[/tex]

    I suppose this makes intuitive sense. But if it's true, it must have a general form. I.e.,

    [tex]\sum_{k=0}^\infty (k+1) r^k = ?[/tex]

    Now, I know that the geometric series converges like so:

    [tex]\sum_{k=0}^\infty r^k = \frac{1}{1-r}[/tex]

    But by multiplying by (k+1) inside the summation completely changes things. Is there a name for this series? Is it true that it converges? If so, what does it converge to?

    This question won't stop plaguing me. Since I don't know what this series is called, I'm having a hard time searching for it on the Internet.

    Thanks!
     
    Last edited: Jan 24, 2010
  2. jcsd
  3. Jan 24, 2010 #2

    jgens

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    Gold Member

    Well, I won't derive the general case for you but I will show you how to evaluate this particular sum (although, I'm not sure that I agree with the answer that someone else provided) . . .

    [tex]S = \sum_{k=1}^{\infty}\frac{k5^k}{6^k} = \frac{5}{6} + 2\left(\frac{5^2}{6^2}\right) + \dots = \frac{5}{6} \left [ 1 + 2\left(\frac{5}{6}\right) + 3\left(\frac{5^2}{6^2}\right) + \dots \right ][/tex]

    Therefore, we know that

    [tex]S = \frac{5}{6} \left [ 1 + 2\left(\frac{5}{6}\right) + 3\left(\frac{5^2}{6^2}\right) + \dots \right ] = \frac{5}{6} \left ( 1 + \sum_{k=1}^{\infty}\frac{(k + 1)5^k}{6^k} \right )[/tex]

    With some simple manipulations, we can put this last sum into a more desirable form

    [tex]\sum_{k=1}^{\infty}\frac{(k + 1)5^k}{6^k} = \sum_{k=1}^{\infty}\frac{k5^k}{6^k} + \sum_{k=1}^{\infty}\left(\frac{5}{6}\right)^k = S + \sum_{k=1}^{\infty}\left(\frac{5}{6}\right)^k[/tex]

    Now, using this expression we find that

    [tex]S = \frac{5}{6} + \frac{5S}{6} + \frac{5}{6}\sum_{k=1}^{\infty}\left(\frac{5}{6}\right)^k[/tex]

    From which it follows that

    [tex]S = 5 + 5\sum_{k=1}^{\infty}\left(\frac{5}{6}\right)^k[/tex]

    Using the formula for the sum of a geometric series, we can evaulate the last sum and find the value of [itex]S[/itex]

    [tex]S = 5 + 5\left(\frac{1}{1 - \frac{5}{6}}\right) = 5 + 5(6) = 35[/tex]

    Edit: Aside from any significant mistakes I may have made in evaluating that particular sum, you should also note that I evaluated the sum using very informal methods.
     
  4. Jan 24, 2010 #3
    Convergence issues aside (actually it converges for every complex r with absolute value less than 1, and converges uniformly on every disk with a radius less than 1)
    A good insight is that it is a derivative of a more simple series:

    [tex]\frac{d}{dr}\sum^{\infty}_{k=0}r^{k+1}=\sum^{\infty}_{k=0}(k+1)r^{k}[/tex]

    And you know that:

    [tex]\sum^{\infty}_{k=0}r^{k+1}=r\sum^{\infty}_{k=0}r^{k}=\frac{r}{1-r}[/tex] (For any r for which it converges)

    And therefore your sum is given by

    [tex]S(r)=\frac{d}{dr}(\frac{r}{1-r})=\frac{(1-r)+r}{(1-r)^{2}}=\frac{1}{(1-r)^{2}}[/tex]

    Which also fits your special case r=5/6.
     
  5. Jan 24, 2010 #4

    jgens

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    Gold Member

    Wow, that's much simpler. It really makes me wish that I was more proficient with infinite series.
     
  6. Jan 24, 2010 #5
    That was just what I was looking for -- not only the solution, but a tool to use in the future :).

    Thank you so much!
     
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