# 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.

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Show that this holds for $p(z)=z(z-1)\cdots(z-k+1)$. – Pierre-Yves Gaillard Dec 7 '11 at 12:25
@Pierre-YvesGaillard I don't see how fixing $p$ helps here, do you mean $q$ ? – Mathmo Dec 7 '11 at 12:30
No, I mean $p$. Write $p(n)=n(n-1)\cdots(n-k+1)$ if you prefer. – Pierre-Yves Gaillard Dec 7 '11 at 12:41
Nice question! +1! And, since you seem to be new here: Welcome to MSE! – Pierre-Yves Gaillard Dec 7 '11 at 16:14

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: $$\sum_{n=0}^\infty\ n^3\ z^n= \frac{z}{(1-z)^2}+ \frac{6\ z^2}{(1-z)^3}+ \frac{6\ z^3}{(1-z)^4}\quad.$$

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I'm afraid I don't follow your argument. Perhaps I haven't covered this area of Maths before - unsure as to how you're defining 'generate' in this context. – Mathmo Dec 7 '11 at 13:18
Dear @TJO: Preliminary question: Do you agree that the set $V\subset\mathbb C[X]$ of those polynomials $p$ for which the statements hold is a vector subspace? – Pierre-Yves Gaillard Dec 7 '11 at 13:34
Gaillaird: I see that it's a subspace, yes. – Mathmo Dec 7 '11 at 13:52
Dear @TJO: Consider the polynomials $X(X-1)\cdots(X-k+1)$ for $k\ge0$. Do you agree that: (1) they generate $\mathbb C[X]$ (as a $\mathbb C$-vector space), and (2) they are in $V$. [(2) requires a small computation. I'd be happy to give more details if necessary.] – Pierre-Yves Gaillard Dec 7 '11 at 14:06
Ah, right, I see how they generate $\mathbb{C}[X]$ . Not sure how this feeds into the question yet though. (Excuse my ignorance) – Mathmo Dec 7 '11 at 14:16

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.

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Thanks! Hadn't even thought of an inductive proof; very useful! – Mathmo Dec 7 '11 at 13:12