Teeny tiny series

Note first that $\sum_{n=0}^\infty x^{n} = \frac{1}{1 - x}$ for $|x| \lt 1$ .

Differentiating both sides, (the LHS term by term), we find that

$\sum_{n=0}^\infty n * x^{n - 1} = \frac{1}{(1 - x)^2}$ .

Multiplying through by $x$ gives us that

$\sum_{n=0}^\infty n * x^{n} = \frac{x}{(1 - x)^2}$ .

Finally, to evaluate the given series we just plug $x = \frac{1}{2}$ into this last equation, giving us a solution of

$\frac{(1/2)}{(1 - (1/2))^2} = \frac{1/2}{1/4} = \boxed{2}$ .