Calculate $\sum_{x=1}^{\infty} \frac{(x-1)}{2^{x}}$

Question

I can prove it converges but I don't know at what value it converges. $\sum_{x=1}^{\infty} \frac{(x-1)}{2^{x}}$

Answer 1

because all the summands are nonnegative you can actually rearrange the sum like this:$$\sum_{x=1}^\infty \frac{x-1}{2^x}= \sum_{k=2}^\infty \underbrace{ \sum_{x=k}^\infty \frac{1}{2^x}}_{=\frac{1}{2^{k-1}}}$$

Answer 2

Hint

$$\sum_{x=1}^\infty (x-1)y^x={d\over dy}\sum_{x=1}^\infty y^{x+1}$$

Answer 3

Start with (for $\lvert u \rvert < 1$ the geometric series converges absolutely, we'll use $u = 1/2$ in the end, so all the manhandling is justified):

$\begin{align*} \sum_{x \ge 1} u^x &= \frac{u}{1 - u} \\ \frac{d}{d u} \sum_{x \ge 1} u^x &= \frac{1}{(1 - u)^2} \\ \sum_{x \ge 1} x u^{x - 1} &= \\ \sum_{x \ge 1} x u^x &= \frac{u}{(1 - u)^2} \\ \sum_{x \ge 1} (x - 1) u^x &= \frac{u}{(1 - u)^2} - \frac{u}{1 - u} \\ &= \frac{u^2}{(1 - u)^2} \end{align*}$

Now replace $u = 1/2$ and get:

$\begin{align*} \sum_{x \ge 1} \frac{x - 1}{2^x} &= \frac{(1/2)^2}{(1 - 1/2)^2} \\ &= 1 \end{align*}$