1. ## Infinite Sums

How can I evaluate the sum from i=0 to infinity of i^2 / (4^i) ?

Also, I would like to check my other answers. For sum i=0-> inf of 1/4^i I got 4/3

For sum i=0-> inf of i/4^i I got 4/9

Sorry this looks kind of crappy.

Let S = $\displaystyle \frac{1^2}{4} + \frac{2^2}{4^2} + \frac{3^2}{4^3} + ...$

We already know that S'(say) = $\displaystyle \frac{1}{4} + \frac{2}{4^2} + \frac{3}{4^3} + ...$ = $\displaystyle \frac{4}{9}$

Observe that S - $\displaystyle \frac{S}{4}$ = $\displaystyle \frac{1^2}{4} + ( \frac{2^2}{4^2} - \frac{1^2}{4^2})+ (\frac{3^2}{4^3} - \frac{1^2}{4^2}) + ...$

$\displaystyle \Rightarrow \frac{3S}{4} = \frac{1^2}{4} + \frac{(2+1)(2-1)}{4^2} + \frac{(3+2)(3-2)}{4^3} + ...$

$\displaystyle \Rightarrow \frac{3S}{4} = \frac{1}{4} + \frac{(2+1)}{4^2} \frac{(3+2)}{4^3} + ...$

$\displaystyle \Rightarrow \frac{3S}{4} = (\frac{1}{4} + \frac{2}{4^2} + \frac{3}{4^3} + ...) + ( \frac{1}{4^2} + \frac{2}{4^3} + \frac{3}{4^4} + ...)$

$\displaystyle \Rightarrow \frac{3S}{4} = S' + \frac{S'}{4} = \frac{5S'}{4}$
$\displaystyle \Rightarrow S = \frac{20}{27}$

Numerators remain the same, only denominators are multiplied by 4. Our goal was to bring the sum into a form that has i instead of $\displaystyle i^2$ in the numerator. We achieved this by using the identity $\displaystyle a^2 - b^2$ = (a+b)(a-b).