<s>[INST] 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
 [/INST] 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 $S=2$. 
</s>