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 November 13th, 2017, 02:15 AM #1 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 104 Thanks: 2 prove question Prove that $\frac{1}{(1-x)^2}=\sum_{n=0}^{\infty}(n+1)x^n$ using term by term differentiation of power series when |x|<1 I need help with this question; here is what I have done so far: $\sum_{n=0}^{\infty}x^n\times\sum_{n=0}^{\infty}x^n =(1+x+x^2+...)^2$ but I am kind of stuck as to where to go from here, thanks! Last edited by skipjack; November 13th, 2017 at 08:30 AM.
 November 13th, 2017, 03:34 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 I don't see how what you "have done so far" has anything to do with this problem. The problem says "using term by term differentiation of power series". Note that $\displaystyle \frac{1}{(1- x)^2}= (1- x)^{-2}$ has anti-derivative $(1- x)^{-1}= \frac{1}{1- x}$. You should be able to recognize that as the sum of the geometric series $\sum_{n=0}^\infty x^n$ which converges uniformly for |x|< 1. Differentiating that "term by term" we have $\sum_{n=1}^\infty nx^{n-1}$ which, letting m= n- 1, is the same as $\sum_{m=0}^\infty (m+1)x^m$. Thanks from greg1313, topsquark and Jaket1
 November 13th, 2017, 08:17 AM #3 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 104 Thanks: 2 Hello. Thanks for the help, but why did you take the anti-derivative? Last edited by skipjack; November 13th, 2017 at 08:32 AM.
 November 13th, 2017, 02:33 PM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Because it gave a geometric series which is easy to sum.

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