I am looking for a step-by-step solution to the following infinite sum:
$\sum_{i=0}^\infty \frac{i^n}{4^i}$
I know that I can find the solution to something like this:
$\sum_{i=0}^\infty \frac{i^2}{4^i}$
by starting with the geometric series and its closed form:
$\sum_{i=0}^\infty x^i = \frac{1}{(1-x)}$
and by differentiating both sides by x then multiplying both sides by x and repeating (perform differentiation by x followed by multiplication by x) i can get:
$\sum_{i=0}^\infty i^2x^i = \frac{x(1+x)}{(1-x)^3}$
at which point I just set x = (1/4) and the RHS evaluates to 20/27.
But I am unsure of what to do when the power of i in the series is set to n, instead of a constant like 2 or 3
Let $s(x, n) =\sum_{k=0}^{\infty} k^n x^k $.
$s(x, 0) =\dfrac1{1-x} $.
$s'(x, n) =\sum_{k=1}^{\infty} k^{n+1} x^{k-1} =\dfrac1{x}\sum_{k=1}^{\infty} k^{n+1} x^{k} =\dfrac1{x}s(x, n+1) $.
Therefore $s(x, n+1) =xs'(x, n) $.
$s(x, 1) =xs'(x, 0) =x(\dfrac1{1-x})' =\dfrac{x}{(1-x)^2} $.
$s(x, 2) =xs'(x, 1) =x(\dfrac{x}{(1-x)^2})' =\dfrac{x(1+x)}{(1-x)^3} $.
$s(x, 3) =xs'(x, 2) =x(\dfrac{x(1+x)}{(1-x)^3})' =\dfrac{x(1+4x+x^2)}{(1-x)^4} $.
$s(x, 4) =xs'(x, 3) =x(\dfrac{x(1+4x+x^2)}{(1-x)^4})' =\dfrac{x (x^3 + 11 x^2 + 11 x + 1)}{(1-x)^5} $.
$s(\frac14, 4) =\sum_{k=0}^{\infty} k^4 (\frac14)^k =\dfrac{\frac14 ((\frac14)^3 + 11 (\frac14)^2 + 11 (\frac14) + 1)}{(1-\frac14)^5} =\dfrac{380}{81} $.