Compute the sum $\sum_{k=1}^{\infty}k^mz^k$ where $|z|<1$ 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).
 A: 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,
$$
\begin{align}
\sum_{k=0}^\infty k^mz^k
&=\sum_{k=0}^\infty\sum_{j=0}^m\stirtwo{m}{j}\binom{k}{j}j!z^k\\
&=\sum_{j=0}^m\stirtwo{m}{j}j!\sum_{k=j}^\infty\binom{k}{j}z^k\\
&=\sum_{j=0}^m\stirtwo{m}{j}j!\sum_{k=j}^\infty\binom{k}{k-j}z^k\\
&=\sum_{j=0}^m\stirtwo{m}{j}j!\sum_{k=j}^\infty(-1)^{k-j}\binom{-j-1}{k-j}z^k\\
&=\sum_{j=0}^m\stirtwo{m}{j}j!\sum_{k=0}^\infty(-1)^{k}\binom{-j-1}{k}z^{k+j}\\
&=\sum_{j=0}^m\stirtwo{m}{j}j!\frac{z^j}{(1-z)^{j+1}}\tag{2}
\end{align}
$$

For $m=2$, $(2)$ gives
$$
\begin{align}
\sum_{k=0}^\infty k^2z^k
&=\frac{z}{(1-z)^2}+2\frac{z^2}{(1-z)^3}\\
&=\frac{z+z^2}{(1-z)^3}\tag{3}
\end{align}
$$
A: 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.
A: 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
$${m\brace q} = m! [z^m] \frac{(\exp(z)-1)^q}{q!}.$$
