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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?

Martin Sleziak
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MT_
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  • Given polynomial $p(n)$ with, say, integer coefficients, we can evaluate the sum of $p(n) x^n,$ using term by term integration or derivative. The radius of convergence will be $1.$ You have several answers now, anyway replace $x$ by $x/8.$ – Will Jagy Mar 04 '14 at 06:42
  • This is a polylogarithm. – Lucian Mar 04 '14 at 08:29

3 Answers3

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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 $p = 1/8$: $$\color{#00f}{\large% \sum_{n = 1}^{\infty}n^{3}\pars{1 \over 8}^{n} = \left.{p - 4p^{2} + p^{3} \over \pars{1 - p}^{4}}\right\vert_{p\ =\ 1/8} = {776 \over 2401}} $$

Felix Marin
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  • What does $p \ni |p| < 1$ mean? Please use standard notation, don't make up your own notation. (And why flood the answer with commands you're not using?) – TMM Mar 04 '14 at 13:39
  • I know this notation. It means "$p$ such that $|p|<1$". But I agree, use such notation only when writing notes for yourself, not when writing for others to read. – GEdgar Mar 04 '14 at 14:42
  • @TMM $\large\ni$ is standard notation. It was widely used when I was an undergraduate. Sometimes ago, I asked to a mathematician about some nice notation that we are loosing. He told me that modern math is making an effort to be more verbose for pedagogical reasons. Any way. I liked it. It is short. Thanks. – Felix Marin Mar 04 '14 at 19:07
  • One of the main problems in many areas of mathematics is that people keep introducing their own notation, instead of following existing notation. There are many ways to write "such that" that are widely accepted and which should not bother anyone, but $\ni$ is not one of them. (At the very least it is ambiguous with $S \ni s$ which means $s \in S$. That alone is already a good reason to abandon $\ni$ for "such that," with so many good alternatives available.) – TMM Mar 04 '14 at 19:18
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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.

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    I get $f'''(1) + 2f''(1) + f'(1) = \sum_{ n = 1}^{\infty} \frac{n^3}{8^n}$ How does one find $f^{(k)} (1)$? Actually I might not have differentiated correctly. When you derive $xf'(x)$, can you just use regular product rule? i.e. $x f''(x) + f'(x)$? – MT_ Mar 04 '14 at 06:51
  • Yes you can use the product rule. But normally I wouldn't compute the derivatives formally like that: I'd actually sum the arithmetic expression to get a rational function, and then compute the actual derivatives of the rational function as I go along. –  Mar 04 '14 at 19:12
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$$\sum {n\choose k}x^n$$ 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.

Gerry Myerson
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