Arithmetic over Geometric

You can see that this series equals: $\frac{1}{2}$ + $\frac{2}{4}$ + $\frac{3}{8}$ + $\frac{4}{16}$ +...

= ( $\frac{1}{2}$ + $\frac{1}{4}$ + $\frac{1}{8}$ + $\frac{1}{16}$ +...)+( $\frac{1}{4}$ + $\frac{1}{8}$ + $\frac{1}{16}$ +...)+( $\frac{1}{8}$ + $\frac{1}{16}$ +...)+...

= $1$ + $\frac{1}{2}$ + $\frac{1}{4}$ + $\frac{1}{8}$ +...

= $2$

$\dfrac{1}{1-x} = \displaystyle \sum_{n=0}^{\infty} x^{n} , |x| < 1$
Differentiating,
$\dfrac{1}{(1-x)^2} = \displaystyle \sum_{n=0}^{\infty} nx^{n-1}$
Multiply by x,
$\displaystyle \sum_{n=0}^{\infty} nx^{n} = \dfrac{x}{(1-x)^2}$
Put $x = \dfrac{1}{2}$
$\displaystyle \sum_{n=0}^{\infty} \dfrac{n}{2^n} = 2$

$\begin{matrix} S & = & \frac { 1 }{ 2 } +\frac { 2 }{ 4 } +\frac { 3 }{ 8 } +\cdots \\ \frac { S }{ 2 } & = & \frac { 1 }{ 4 } +\frac { 2 }{ 8 } +\frac { 3 }{ 16 } +\cdots \\ S-\frac { S }{ 2 } & = & \frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 8 } +\cdots \end{matrix}$

This is a Geometric Progression. Using the formula,

$\begin{matrix} S-\frac { S }{ 2 } & = & \frac { \frac { 1 }{ 2 } }{ 1-\frac { 1 }{ 2 } } \\ \frac { S }{ 2 } & = & 1 \\ S & = & 2 \end{matrix}$