Evaluating infinite series

Main Question or Discussion Point

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)?
 
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Answers and Replies

Galileo
Science Advisor
Homework Helper
1,989
6
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.
 
663
0
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:
 
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Thanks for all the help :D
 
bbailey
Gold Member
2
0
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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