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How would I go about finding the sum of this series? $$\sum_{n=2}^\infty \frac{n(n-1)}{2^{n-2}}$$

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Try (really: try ) the n-th root test... –  DonAntonio Mar 31 '13 at 9:16
    
Do you mean $\sum_{n=2}^\infty \frac{n(n-1)}{2^{n-2}}$ ? –  Santosh Linkha Mar 31 '13 at 9:19
    
@experimentX yeah that's what i mean –  Billy Thompson Mar 31 '13 at 9:20
    
differentiate and put $x=\frac{1}{2}$ down below. –  Santosh Linkha Mar 31 '13 at 9:21

1 Answer 1

Hint: for $|x|<1$, $$\sum_{n=0}^\infty x^n = \frac{1}{1-x},$$ so $$\sum_{n=1}^\infty nx^{n-1}=???$$ and $$\sum_{n=2}^\infty n(n-1)x^{n-2}=???$$


Differentiate twice to get $$\sum_{n=2}^\infty n(n-1)x^{n-2}=\frac{2}{(1-x)^3},$$ and put $x=\frac{1}{2}$.

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(+1): Added to one of my most loved answers. –  Parth Kohli Mar 31 '13 at 9:26
    
I"m sorry, I don't understand how this helps edit: nvm thanks a lot –  Billy Thompson Mar 31 '13 at 9:56
    
To get it clear: Since the geometric series is absolutely convergent (if |x|<1), one can change the order of differentiation and summation? –  gofvonx Mar 31 '13 at 10:04
    
@LeoSti: There are various ways to look at this, but yes, it's because a power series is uniformly convergent on $[-r+\varepsilon,r-\varepsilon]$ for any $\varepsilon>0$, where $r$ is the radius of convergence. See Theorem 8.1 in Baby Rudin. –  wj32 Mar 31 '13 at 10:07

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