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

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