# Find the value of $\sum_{k=1}^{\infty} \frac{k}{2^k}$ [duplicate]

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

I know that $\sum_{k=1}^{\infty} \frac{1}{2^k}=1$ so the answer is $\frac{k(k+1)}{2}*1$?

• The result could not depend on $k$! Have you tried wolfram alfa? Mar 20, 2016 at 17:14
• You cannot multiply them together like that Mar 20, 2016 at 17:14
• Also, $$\sum_{k=1}^{\infty} k \neq \frac{k(k+1)}{2}$$ Mar 20, 2016 at 17:15

Hint: for $|x|<1$, $$\sum \limits_{k=1}^{\infty} x^n=\frac{1}{1-x}.$$
Try differentiating both sides of this equation and let $x= \frac{1}{2}$.
Here is a useful finite evaluation: $$1+r+r^2+...+r^n=\frac{1-r^{n+1}}{1-r}, \quad |r|<1. \tag1$$ Then by differentiating $(1)$ you get $$1+2r+3r^2+...+nr^{n-1}=\frac{1-r^{n+1}}{(1-r)^2}+\frac{-(n+1)r^{n}}{1-r}, \quad |r|<1, \tag2$$ and by making $n \to +\infty$ in $(2)$, using $|r|<1$, gives
$$1+2r+3r^2+...+nr^{n-1}+...=\frac{1}{(1-r)^2} \tag3$$
If you multiply $(3)$ by $r$ and set $r=\dfrac12$, you obtain an answer to your question.