Solving for $\sum_{n = 1}^{\infty} \frac{n^3}{8^n}$?

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With $p\ \ni\ \verts{p} < 1$: \begin{align} \sum_{n = 1}^{\infty}p^{n} &= {p \over 1 - p} = - 1 + {1 \over 1 - p} \end{align} Derive respect of $p$ and after that multiply by $p$: \begin{align} \sum_{n = 1}^{\infty}np^{n} &= {p \over \pars{1 - p}^{2}} \\ \sum_{n = 1}^{\infty}n^{2}p^{n} &= -\,{p + p^{2} \over \pars{1 - p}^{3}} \\ \sum_{n = 1}^{\infty}n^{3}p^{n} &= {p - 4p^{2} + p^{3} \over \pars{1 - p}^{4}} \end{align}

Set $p = 1/8$: $$\color{#00f}{\large% \sum_{n = 1}^{\infty}n^{3}\pars{1 \over 8}^{n} = \left.{p - 4p^{2} + p^{3} \over \pars{1 - p}^{4}}\right\vert_{p\ =\ 1/8} = {776 \over 2401}} $$

There turns out to be a standard trick that applies here: let

$$ f(x) = \sum_{i=1}^{\infty} \frac{x^n}{8^n} $$

We can compute this sum because it is a geometric series. The neat idea, now, is that

$$ f'(x) = \sum_{i=1}^{\infty} \frac{n x^{n-1}}{8^n} $$

or alternatively,

$$ x f'(x) = \sum_{i=1}^{\infty} \frac{n x^n}{8^n} $$

Repeat a few times, then plug in $x=1$, and you get the answer.

$$\sum {n\choose k}x^n$$ is easy to sum, using the binomial theorem. Then if you can express $n^k$ in terms of ${n\choose0},{n\choose1},\dots,{n\choose k}$, you can get a formula for $\sum n^kx^n$. Expressing powers of $n$ in terms of those binomial coefficients can be done using Stirling numbers, which I invite you to look up.