TITLE: What does $\sum_{k=0}^\infty \frac{k}{2^k}$ converge to?
QUESTION [17 upvotes]: 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?

REPLY [9 votes]: You can also view it this way, which is quite intuitive, though possibly not that rigorously:
\begin{align*}
    \sum_{k=0}^\infty \frac{k}{2^k}
    &= \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \frac{5}{32} + \cdots \\
    &= \frac12 + \Bigl(\frac14 + \frac14\Bigr) + \Bigl(\frac18 + \frac18 + \frac18\Bigr) + \Bigl(\frac1{16} + \frac1{16} + \frac1{16} + \frac1{16}\Bigr) + \cdots \\
    &= \Bigl(\frac12 + \frac14 + \frac18 + \frac1{16} + \cdots\Bigr) + \Bigl(\frac14 + \frac18 + \frac1{16} + \cdots\Bigr) + \Bigl(\frac18 + \frac1{16} + \cdots\Bigr) + \cdots \\
    &= 1 + \frac12 + \frac14 + \frac18 + \cdots \\
    &= 2.
\end{align*}