# Show that $\sum_{n = 1}^{+\infty} \frac{n}{2^n} = 2$ [duplicate]

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

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