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

I have no idea to calculate this sum. I　try shift index since $k=0$ gives 0, tis won't change the sum. But I don't know how to keep going. Can someone give me a hint or suggestion to calculate this sum? Thanks
 A: Hint. Probably you know that $x\not=1$,
$$\sum_{k=0}^{N-1}x^k=\frac{1-x^{N}}{1-x}.$$
What happens if you differentiate both sides?
A: Hint. Fubini.
$$k=\sum_{i=1}^k1$$
A: A solution without calculus:
Separate the sum into
$$\sum_{k=1}^{\infty} \frac{1}{2^k}\sum_{n=1}^k 1$$
Interchanging summation:
$$\sum_{n=1}^{\infty}\sum_{k=n}^{\infty} \frac{1}{2^k}$$
Can you solve it from here?
A: Split it up into two summations:
$$\sum_{k=1}^{\infty}\frac{k}{2^k}=\sum_{i=1}^{\infty}\sum_{j=i}^{\infty}\frac{1}{2^j}$$
Trying out simple cases, you find the total sum is:
$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+... = 1$ for $i = 1$
$\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+... = \frac{1}{2}$ for $i = 2$
Etc.
Then you have $1 + \frac{1}{2} + \frac{1}{4} + ... = 2$, as desired.
A: Here's another solution. 
Let 
\begin{align}
S=\sum^\infty_{k=1}\frac{k}{2^k} =&\  \sum^\infty_{k=1} \frac{k-1+1}{2^k} = \frac{1}{2}\sum^\infty_{k=1} \frac{k-1}{2^{k-1}} + \sum^\infty_{k=1} \frac{1}{2^k}\\
=&\ \frac{1}{2}\sum^\infty_{k=2} \frac{k-1}{2^{k-1}}+\frac{1}{2} = \frac{1}{2}\sum^\infty_{k=1}\frac{k}{2^k}+1 \\
=& \frac{1}{2}S+1. 
\end{align}
Solving for $S$ yields $S=2$. 
A: Let 
$$ S = \sum_{k=1}^\infty \frac{k}{2^k} = \frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots$$.
Then
$$ 2S = \frac{1}{2^0}+\frac{2}{2^1}+\frac{3}{2^2}+\cdots$$
and 
$$S = 2S-S = \frac{1}{2^0} + \frac{1}{2^1}+\frac{1}{2^2} + \cdots = 2$$
