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?

Solution 1:

$$S=\sum_{k=0}^\infty{k\over2^k}=2\sum_{k=0}^\infty{k\over2^{k+1}}=2\sum_{k=1}^\infty{k-1\over2^k}=2S-2\sum_{k=1}^\infty{1\over2^k}=2S-2\\ \therefore S=2$$

To prove convergence of $S$ we can use ratio test...

Solution 2:

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 $x=\frac{1}{2}$.

Solution 3:

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*}