I have the series $\sum_{n=0}^\infty \frac{n}{2^n}$. I must show that it converges to 2.   I was given a hint to take the derivative of $\sum_{n=0}^\infty x^n$ and multiply by $x$ , which gives   $\sum_{n=1}^\infty nx^n$ , or $\sum_{n=0}^\infty nx^n$.   Clearly if I take $x=\frac{1}{2}$ , the series is  $\sum_{n=0}^\infty \frac{n}{2^n}$. How do I proceed from here?
This problem comes from another equation on another question (. I tried to split it in half but I found out that $$\sum_{k=0}^\infty \frac{k}{2^k}$$ can't be divided. Knowing that $$\sum_{k=0}^\infty x^k=\frac{1}{1-x}$$ I wrote that $$\sum_{k=0}^\infty \frac{k}{2^k}=\sum_{k=0}^\infty \left(\frac{\sqrt[k] k}{2}\right)^k=\frac{1}{1-\frac{\sqrt k}{2}}=\frac{2}{2-\sqrt[k] k}$$ But that's not what I wanted. Could anyone help me?
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