Summation differentiation geometric series

  • #1

Homework Statement


find the sum for

[tex]\sum_{k=1}^{\infty} kx^{k} [/tex]


Homework Equations



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

The Attempt at a Solution



[tex] \sum_{k=1}^{\infty} kx^{k} = \sum_{n=0}^{\infty}(n+1)x^{n+1} = x\sum_{n=0}^{\infty} (n+1)x^{n} = x \frac{d}{dx} \sum_{n=0}^{\infty}x^{n+1}[/tex]

I'm not sure how to proceed; thoughts?
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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I think from here

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


You should just differentiate both sides w.r.t. x and then change the index.
 
  • #3
does it matter that it's xn+1 and not xn?
 
  • #4
tiny-tim
Science Advisor
Homework Helper
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Hi Somefantastik! :smile:

(ooh, you just changed it! :biggrin:)

That's fine … now you need to get that ∑xn+1 to be part of a ∑xn starting at 0 (so you may need to subtract something :wink:).
 
  • #5
How about

[tex] x \frac{d}{dx} \sum_{n=0}^{\infty}x^{n+1} = x \frac{d}{dx}\left[x\left(\sum_{n=0}^{\infty}x^{n}\right)\right] = x \frac{d}{dx}\left[x\frac{1}{1-x}\right][/tex]
 
  • #6
tiny-tim
Science Advisor
Homework Helper
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Yes, that's fine also. :smile:

( x/(1-x) = 1/(1-x) - 1 )

(though in the form 1/(1-x) - 1, it's easier to differentiate! :wink:)
 

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