Infinite series in terms of x

  • #1

Homework Statement



Hi,

How do i determine de result in terms of x of this series for x < 1:

(Sum(i=0..+infinity; i*x^i))/(Sum(i=0..+infinity;x^i)

Thanks


The Attempt at a Solution



I know that (Sum(i=0..+infinity;x^i) will tend do 1/(1-x) but i don't know what the numerator will tend to

Thanks
 
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  • #2
Hi benf.stokes! :smile:

(have a sigma: ∑ and an infinity: ∞ and try using the X2 and X2 tags just above the Reply box :wink:)
benf.stokes said:
(Sum(i=0..+infinity; i*x^i))/(Sum(i=0..+infinity;x^i)

I know that (Sum(i=0..+infinity;x^i) will tend do 1/(1-x) but i don't know what the numerator will tend to

Thanks

Hint: integrate. :smile:
 
  • #3
Thanks, I figured it out but by differentiating (sorry for the delay but I was netless for a few days):

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

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

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

How would it be done by integrating? The other way around?
 
Last edited:
  • #4
Sum the integral, and then differentiate that sum. :smile:
 

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