What does $\sum_{k=0}^\infty \frac{k}{2^k}$ converge to? This problem comes from another equation on another question (this one).
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?

Start with:
$$\frac{1}{1-x}=\sum_{k=0}^{\infty}x^k.$$
Then take derivative with respect to $x$.
$$\frac{1}{(1-x)^2}=\sum_{k=1}^{\infty}kx^{k-1}.$$
Multiply by $x$.
$$\frac{x}{(1-x)^2}=\sum_{k=1}^{\infty}kx^{k}.$$
Now substitute $x=\frac{1}{2}$.
