Evaluate the Sum $\sum_{i=0}^\infty \frac {i^N} {4^i}$ [duplicate]

Solution 1:

Note that for $\vert x \vert < 1$, we have $$\sum_{i=0}^{\infty} x^i = \dfrac1{1-x}$$ We then have $$\sum_{i=0}^{\infty} ix^i = x \dfrac{d}{dx}\left(\sum_{i=0}^{\infty} x^i \right) = x \dfrac{d}{dx}\left(\dfrac1{1-x}\right)$$ If we denote the operator $x\dfrac{d}{dx}$ as $L$, we then have $$\sum_{i=0}^{\infty} ix^i = L\left(\dfrac1{1-x}\right)$$ $$\sum_{i=0}^{\infty} i^2x^i = L\left(L\left(\dfrac1{1-x}\right) \right) = L^2\left(\dfrac1{1-x}\right)$$ And in general, $$\sum_{i=0}^{\infty} i^Nx^i = L^N\left(\dfrac1{1-x}\right)$$ Hence, $$\sum_{i=0}^{\infty} \dfrac{i^N}{4^i} = \left.L^N\left(\dfrac1{1-x}\right) \right \vert_{x=1/4}$$ Also, this is nothing but $$\text{Li}_{-N}\left(\dfrac14\right),$$ where $\text{Li}_s(z) = \displaystyle \sum_{k=1}^{\infty} \dfrac{z^k}{k^s}$, the Polylogarithm function.

Solution 2:

Related techniques:(I), (II), (III). Recalling the identity for the operator $(xD)^n$, where $D=\frac{d}{dx}$,

$$ (xD)^n = \sum_{k=0}^{n}\begin{Bmatrix} n\\k \end{Bmatrix}x^k D^k , $$

where $\begin{Bmatrix} n\\k \end{Bmatrix}$ are the Stirling numbers of the second kind. One can get a closed form for the sum in terms of a finite sum. Applying this operator to the function $\frac{1}{1-x}$

$$ \sum_{k=0}^{\infty}{k^n}{x^k}= (xD)^n\frac{1}{1-x} = \sum_{k=0}^{n}\begin{Bmatrix} n\\k \end{Bmatrix}x^k D^k \frac{1}{1-x} $$

$$\implies \sum_{k=0}^{\infty}{k^n}{x^k}= \sum_{k=0}^{n}\begin{Bmatrix} n\\k \end{Bmatrix} k!\frac{x^k}{(1-x)^{k+1}}. $$

Subs $x=\frac{1}{4}$ in the above identity gives

$$ \sum_{k=0}^{\infty}\frac{k^n}{4^k} = 4\sum_{k=0}^{n}\begin{Bmatrix} n\\k \end{Bmatrix} \frac{k!}{3^{k+1}}. $$

Here is some values of the last formula for $1\leq n \leq 7$

$$ \left\{ \frac{4}{9},{\frac {20}{27}},{\frac {44}{27}},{\frac {380}{81}},{ \frac {4108}{243}},{\frac {17780}{243}},{\frac {269348}{729}} \right\} . $$