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?
 A: $$S=\sum_{k=0}^\infty{k\over2^k}=2\sum_{k=0}^\infty{k\over2^{k+1}}=2\sum_{k=1}^\infty{k-1\over2^k}=2S-2\sum_{k=1}^\infty{1\over2^k}=2S-2\\$$
Using the ratio test, we show that $S$ converges. Therefore, we are able to rearrange the above result to solve for $S$ which gives us the desired result of $S=2$.
A: 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 $x=\frac{1}{2}$.
A: You can also view it this way, which is quite intuitive.
\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*}
We are able to rearrange the terms of the infinite series in this way because the series is absolutely convergent. It is important that we meet this condition, as without this we would not be able to reorder the terms of the series and group them in this way.
