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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. Tags prove, question Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post 450081592 Applied Math 1 September 15th, 2010 03:35 AM 450081592 Calculus 0 September 14th, 2010 05:26 PM 450081592 Linear Algebra 1 February 25th, 2010 02:16 AM igalep132 Calculus 8 February 7th, 2008 04:36 AM jiasyuen Algebra 0 December 31st, 1969 04:00 PM

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