Show that $\sum_{n = 1}^{+\infty} \frac{n}{2^n} = 2$. I have no idea to solve this problem. Anyone could help me?

  • $\begingroup$ Take a geometric series, and differentiate it. $\endgroup$ – Daniel Fischer Aug 3 '13 at 15:24
  • $\begingroup$ Why should this question be re-opened? (It seems to be a genuine duplicate to me...) $\endgroup$ – user1729 Aug 3 '13 at 17:45

Consider the following

$$\frac{1}{1-x} = \sum_{k\geq 0}x^ k $$

The series converges for $|x|<1$

Differentiating both sides we have

$$\frac{1}{(1-x)^2} = \sum_{k\geq 1}k x^{k-1} $$

$$\frac{x}{(1-x)^2} = \sum_{k\geq 1}k x^{k} $$

Now put $x=\frac{1}{2}$

$$2 = \sum_{k\geq 1}\frac{k}{2^k} $$


Consider $S_{1}=\sum_{n=1}^{\infty} 1/2^{n}=1$
$S_{2}=\sum_{n=2}^{\infty} 1/2^{n}=1/2$

$S_{k}=\sum_{n=k}^{\infty} 1/2^{n}=1/2^{k-1}$
Then the sum that you are taking is equivalent to $\sum_{k=1}^{\infty}S_{k}$.
This is an infinite geometric series with 1st term equal to 1 and common ratio 1/2, thus the sum is 2.


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