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November 13th, 2017, 03:15 AM   #1
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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 09:30 AM.
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November 13th, 2017, 04:34 AM   #2
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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 $\sum_{m=0}^\infty (m+1)x^m$.
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November 13th, 2017, 09:17 AM   #3
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Hello. Thanks for the help, but why did you take the anti-derivative?

Last edited by skipjack; November 13th, 2017 at 09:32 AM.
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November 13th, 2017, 03:33 PM   #4
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Because it gave a geometric series which is easy to sum.
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