$\displaystyle \sum_{x=0}^{\infty}\frac{x}{2^x}$
All I know is that it converges. Really have nothing more to say.
Hello, lausing!
$\displaystyle S \;=\;\sum^{\infty}_{x=1} \frac{x}{2^x}$
$\displaystyle \begin{array}{cccccc}\text{We have:} & S &=& \dfrac{1}{2} + \dfrac{2}{2^2} + \dfrac{3}{2^3} + \dfrac{4}{2^4} + \hdots \\ \\[-3mm]
\text{Multiply by }\frac{1}{2}\!: & \frac{1}{2}S &=& \quad\;\;\dfrac{1}{2^2} + \dfrac{2}{2^3} + \dfrac{3}{2^4} + \hdots \end{array}$
. . $\displaystyle \text{Subtract: }\;\;\tfrac{1}{2}S \;\;=\;\;\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \hdots $ .[1]
The right side is a geometric series with: .$\displaystyle a = \tfrac{1}{2},\;\;r = \tfrac{1}{2}$
. . Its sum is: .$\displaystyle \frac{\frac{1}{2}}{1-\frac{1}{2}} \:=\:1$
Hence [1] becomes: .$\displaystyle \tfrac{1}{2}S \;=\;1$
Therefore: .$\displaystyle S \;=\;2$