Solving for $\sum_{n = 1}^{\infty} \frac{n^3}{8^n}$? I was trying to solve $ \displaystyle \sum_{n = 1}^{\infty} \frac{n^3}{8^n}$ and I found a way to solve it and I want if there are generalizations for, say, $\displaystyle \sum_{n=1}^{\infty} \frac{n^k}{a^n}$ in terms of $k$ and $a$. I would also like to know if there is  a better way to solve it. Here's how I did it:
First I decomposed the series into the following sums:
$S_1  = \frac{1}{8} + \frac{1}{64} + \dots = \frac{\frac{1}{8}}{\frac{7}{8}}$
$S_2 = \frac{7}{64} + \frac{7}{512} + \dots = \frac{\frac{7}{64}}{\frac{7}{8}}$
$S_3 = \frac{19}{512} + \frac{19}{4096} + \dots = \frac{\frac{19}{512}}{\frac{7}{8}}$
And deduced that the sum can be written as $\frac{8}{7} \displaystyle \sum_{n = 1}^{\infty} \frac{3n^2 - 3n + 1}{8^n}$
$\displaystyle \sum_{n = 1}^{\infty} \frac{1}{8^n}$ is easy to evaluate -- it's $\frac{1}{7} $by geometric series
$\displaystyle \sum_{n = 1}^{\infty} \frac{n}{8^n}$ can be evaluated in a whole host of ways to get an answer of $\frac{8}{49}$.
It remains to evaluate $\displaystyle \sum_{n = 1}^{\infty} \frac{n^2}{8^n}$, for which I took a similar approach as the cubics by decomposing it into many sums:
$T_1 = \frac{1}{8} + \frac{1}{64} + \dots = \frac{\frac{1}{8}}{\frac{7}{8}}$
$T_2 = \frac{3}{64} + \frac{3}{512} + \dots = \frac{\frac{3}{64}}{\frac{7}{8}}$
And so forth, coming to the conclusion that it is equal to $\frac{8}{7} \displaystyle \sum_{n = 1}^{\infty} \frac{2n-1}{8^n}$
Now, I used this information and the above values for $\displaystyle \sum_{n = 1}^{\infty} \frac{1}{8^n}$ and $\displaystyle \sum_{n = 1}^{\infty} \frac{n}{8^n}$ to get the sum as $\frac{776}{2401}$, which is confirmed by WA.
So, I would like to reiterate here: Is there a simpler way to compute this sum, and are there any known generalizations for this problem given an arbitrary $a$ in the denominator and arbitrary $k$ as the exponent in the numerator?
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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 $p = 1/8$:
  $$\color{#00f}{\large%
\sum_{n = 1}^{\infty}n^{3}\pars{1 \over 8}^{n} =
\left.{p - 4p^{2} + p^{3} \over \pars{1 - p}^{4}}\right\vert_{p\ =\ 1/8}
= {776 \over 2401}}
$$

A: 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.
A: $$\sum {n\choose k}x^n$$ 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. 
