Compute the sum $\sum_{k=1}^{\infty}k^mz^k$ where $|z|<1$

Question

Do you know how to find the limit of $\sum_{k=1}^{\infty}k^mz^k$ where $|z|<1$ and m is a natural number? I've tried to google it in wiki but I do not understand the closed form (http://en.wikipedia.org/wiki/List_of_mathematical_series).

Answer 1

Using Stirling Numbers of the Second Kind, we have $\newcommand{\stirtwo}[2]{\left\{{#1}\atop{#2}\right\}}$ $$ k^m=\sum_{j=0}^m\stirtwo{m}{j}\binom{k}{j}j!\tag{1} $$ Therefore, $$ \begin{align} \sum_{k=0}^\infty k^mz^k &=\sum_{k=0}^\infty\sum_{j=0}^m\stirtwo{m}{j}\binom{k}{j}j!z^k\\ &=\sum_{j=0}^m\stirtwo{m}{j}j!\sum_{k=j}^\infty\binom{k}{j}z^k\\ &=\sum_{j=0}^m\stirtwo{m}{j}j!\sum_{k=j}^\infty\binom{k}{k-j}z^k\\ &=\sum_{j=0}^m\stirtwo{m}{j}j!\sum_{k=j}^\infty(-1)^{k-j}\binom{-j-1}{k-j}z^k\\ &=\sum_{j=0}^m\stirtwo{m}{j}j!\sum_{k=0}^\infty(-1)^{k}\binom{-j-1}{k}z^{k+j}\\ &=\sum_{j=0}^m\stirtwo{m}{j}j!\frac{z^j}{(1-z)^{j+1}}\tag{2} \end{align} $$

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For $m=2$, $(2)$ gives $$ \begin{align} \sum_{k=0}^\infty k^2z^k &=\frac{z}{(1-z)^2}+2\frac{z^2}{(1-z)^3}\\ &=\frac{z+z^2}{(1-z)^3}\tag{3} \end{align} $$

Answer 2

Start from: $$ \sum_{k\geq 1}z^k = \frac{z}{1-z}$$ then apply the operator $\varphi: f(z)\to z\cdot f'(z) $ to both terms $m$ times.

That gives that for any $|z|<1$ we have: $$ \sum_{k\geq 1}k\,z^k = \frac{z}{(1-z)^2},\quad \sum_{k\geq 1}k^2 z^k=\frac{z+z^2}{(1-z)^3}$$ and so on.

Answer 3

Suppose we seek to evaluate $$\sum_{k\ge 1} k^m z^k.$$

Put $$k^m = \frac{m!}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{m+1}} \exp(kw) \; dw \\ = \frac{m!}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{m+1}} \sum_{q=0}^k {k\choose q} (\exp(w)-1)^q \; dw \\ = \frac{m!}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{m+1}} \sum_{q=0}^k \frac{k!}{(k-q)!} \frac{(\exp(w)-1)^q}{q!} \; dw \\ = \sum_{q=0}^k \frac{k!}{(k-q)!} \frac{m!}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{m+1}} \frac{(\exp(w)-1)^q}{q!} \; dw \\ = \sum_{q=0}^k \frac{k!}{(k-q)!} {m\brace q}.$$

This yields for the sum $$\sum_{k\ge 0} z^k \sum_{q=0}^k \frac{k!}{(k-q)!} {m\brace q}.$$ where we have included $k=0$ since ${m\brace 0} = 0.$

This becomes $$\sum_{q\ge 0} {m\brace q} \sum_{k\ge q} \frac{k!}{(k-q)!} z^k = \sum_{q\ge 0} {m\brace q} \sum_{k\ge 0} \frac{(k+q)!}{k!} z^{k+q} \\ = \sum_{q\ge 0} {m\brace q} (q!\times z^q) \sum_{k\ge 0} {k+q\choose k} z^{k} = \sum_{q=0}^m {m\brace q} \frac{q!\times z^q}{(1-z)^{q+1}}.$$

Remark. Here we have used the fact that the combinatorial species of set partitions with sets marked is $$\mathfrak{P}(\mathcal{U}(\mathfrak{P}_{\ge 1}(\mathcal{Z})))$$ which gives the generating function $$G(z, u) = \exp(u(\exp(z)-1))$$ so that $${m\brace q} = m! [z^m] \frac{(\exp(z)-1)^q}{q!}.$$