Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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 $x=\frac{1}{2}$.

share|cite|improve this answer
(+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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.