Complex Analysis: Radius of convergence of Power Series Let p be a polynomial of degree $k>0$. Prove that $\sum p(n)z^n$ has radius of convergence $1$ and that there exists a polynomial $q(z)$ of degree $k$ such that $$\sum_{n=0}^{\infty} p(n) z^n=q(z)(1-z)^{-(k+1)}, \qquad  (|z|<1)$$
I've shown the radius of convergence is $1$; not sure how to apporach the second part.
 A: Both statements follow immediately from the fact that the polynomials 
$$
X\ (X-1)\ \cdots\ (X-k+1),\quad k\ge0,
$$
generate $\mathbb C[X]$ as a complex vector space.
EDIT 1. In fact we have 
$$
\sum_{n=0}^\infty\ p(n)\ z^n=\sum_{j=0}^\infty\ (\Delta^j p)(0)\ \frac{z^j}{(1-z)^{j+1}}\quad,
$$
where $\Delta$ is defined by 
$$
(\Delta p)(X):=p(X+1)-p(X).
$$
EDIT 2. I'll try to give a detailed explanation:
For $j\ge0$ put
$$
p_j:=X\ (X-1)\ \cdots\ (X-j+1)
$$
Let $V$ be the set of those polynomials $p\in\mathbb C[X]$ for which the statements hold
Clearly, 


*

*$V\subset\mathbb C[X]$ is a vector subspace, 

*the $p_j$ generate $\mathbb C[X]$ (as a $\mathbb C$-vector space). 
Thus it suffices to show that $p_j$ is in $V$ for all $j$. But we have
$$
\sum_{n=0}^\infty\ p_j(n)\ z^n
=\sum_{n=0}^\infty\ n\ (n-1)\ \cdots\ (n-j+1)\ z^n
$$
$$
=z^j\ \sum_{n=0}^\infty\ \frac{d^j}{dz^j}\ z^n
=z^j\ \frac{d^j}{dz^j}\ (1-z)^{-1}
=\frac{j!\ z^j}{(1-z)^{j+1}}\quad.
$$
If we finally observe that $\Delta\ p_n=n\ p_{n-1}$, we get the all the statements in the question and in the previous edit.
EDIT 3. Here a numerical illustration: We have 
$$
\sum_{n=0}^\infty\ n^3\ z^n=
\frac{a}{1-z}+
\frac{b\ z}{(1-z)^2}+
\frac{c\ z^2}{(1-z)^3}+
\frac{d\ z^3}{(1-z)^4}\quad.
$$
To compute $a,b,c,d$ we form the array
$$
\begin{matrix}
0&1&8&27\\ \\
1&7&19\\ \\
6&12\\ \\
6
\end{matrix}
$$
as follows. In the first row we write the cubes of $0,1,2,3$. In the second row we put the differences between consecutive entries of the first row: $1-0=1$, $8-1=7$, $27-8=19$, and we form the third and fourth row in the same way: $7-1=6$, $19-7=12$, $12-6=6$. Then the our numbers $a,b,c,d$ are equal, in this order, to the numbers $0,1,6,6$ appearing in the first column:
$$
\sum_{n=0}^\infty\ n^3\ z^n=
\frac{z}{(1-z)^2}+
\frac{6\ z^2}{(1-z)^3}+
\frac{6\ z^3}{(1-z)^4}\quad.
$$
A: For the first part: Consider sums of the form $ \sum n^k z^n .$ If $ |z|>1 $ then the terms do not tend to $0$ so the sum diverges. If $ |z|<1 $ then the ratio of consecutive terms has magnitude $$ \frac{ (n+1)^k |z|^{n+1} }{n^k |z|^n} = \left( 1+ \frac{1}{n} \right)^k |z| \to |z|<1 $$ as $n\to \infty.$ Thus, by the ratio test the series converges and we conclude the radius of convergence of this sum is $1.$ Since $ \sum p(n) z^n $ is a sum of series of that form, it also has radius of convergence $1.$
For the second part: Proceed by induction on the degree. The base case is simple. Assume it holds for all polynomials up to degree $k.$ Then let $ h(n) = an^{k+1} + t(n) $ where $a\neq 0$ and $t(n)$ is a polynomial with degree $k.$ 
Then $$ \sum h(n)z^n = a \sum n^{k+1} z^n + \sum t(n) z^n = a\sum n^{k+1} z^n + g(z)(1-z)^{-(k+1) } $$
where $g$ is a polynomial of degree $k$ (the second equality is by the induction hypothesis).
Also by the induction hypothesis, $\sum n^k z^n = d(n)(1-z)^{-(k+1)} $ for some degree $k$ polynomial $d.$ Differentiating both sides gives $ \sum n^{k+1} z^{n-1} = d'(n) (1-z)^{-(k+1)} + d(n) (k+1) (1-z)^{-(k+2)}.$
Thus, $$\sum h(n) z^n = az(d'(n) (1-z)^{-(k+1)} + d(n) (k+1) (1-z)^{-(k+2)}) + g(z) (1-z)^{-(k+1)}$$ $$ = \left( az\cdot d'(n)(1-z)  + (k+1)a \cdot d(n)z  + g(z) (1-z) \right)(1-z)^{-(k+2)}$$
 $$ = j(n) (1-z)^{-(k+2)} $$
where $j$ is a degree $k+1$ polynomial, which proves your statement by induction. 
