November 13th, 2017, 02:15 AM  #1 
Member Joined: Jan 2016 From: Blackpool Posts: 95 Thanks: 2  prove question
Prove that \[\frac{1}{(1x)^2}=\sum_{n=0}^{\infty}(n+1)x^n\] 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: 3,248 Thanks: 887 
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 antiderivative $(1 x)^{1}= \frac{1}{1 x}$. 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 $\sum_{n=1}^\infty nx^{n1}$ which, letting m= n 1, is the same as $\sum_{m=0}^\infty (m+1)x^m$. 
November 13th, 2017, 08:17 AM  #3 
Member Joined: Jan 2016 From: Blackpool Posts: 95 Thanks: 2 
Hello. Thanks for the help, but why did you take the antiderivative?
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: 3,248 Thanks: 887  Because it gave a geometric series which is easy to sum.


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