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Evaluating infinite series

  1. Aug 28, 2005 #1
    I am having trouble evaluating the sum [tex]\sum_{i=1}^{\infty}\frac{i}{4^i}[/tex] by hand.

    My TI-89 is giving me an answer of 4/9 or 0.44 repeating, but I am uncertain how to go about solving this by hand and proving the calculator's result. To my knowledge, no identity or easy quick fix like the Integral test exists. I tried solving for the sum S by messing around with the first 5 or 6 terms, but it was all to no avail. Can anyone here point me in the direction towards evaluating this by hand in terms of a method(not asking anyone to do the problem itself for me)?
     
    Last edited: Aug 28, 2005
  2. jcsd
  3. Aug 28, 2005 #2

    Galileo

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    Do you know the geometric series?

    [tex]\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}[/tex]
    if |x|<1.

    You can make this look like your sum by differentiation.
     
  4. Aug 28, 2005 #3
    I will provide a general case, you can figure out the rest:

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

    Now taking the derivative of both sides gives the following:

    [tex]\sum_{n=0}^{\infty}nx^{n-1}=\frac{1}{(1-x)^2}[/tex]

    Multiply both sides by x:

    [tex]\sum_{n=0}^{\infty}nx^n=\frac{x}{(1-x)^2}[/tex]

    You can apply this to your problem by changing the lower indices and changing x to (1/4).

    I hope this helps,

    Alex

    Edit: It looks like someone got to it just before I submitted this :smile:
     
    Last edited: Aug 28, 2005
  5. Aug 28, 2005 #4
    Thanks for all the help :D
     
  6. Oct 30, 2011 #5

    bbailey

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    Gold Member

    I am trying to find the sum of the infinite series i=1 to infinity, of (e^n)/(3^(n-1)). I know it converges, but I am struggling trying to find the limit. can someone help me.
     
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