# Evaluate: $\sum_{n=2}^\infty \frac{n(n-1)}{2^{n-2}}$

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

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 $x=\frac{1}{2}$.
@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