polynomial, power series, radius of convergnce

Let $p(x)$ be a polynomial of degree $N$ then the radius of convergence of the power series $$\sum_{n=0}^{\infty}p(n)x^n$$

1. depends on $N$

2. is $1$ for all $N$

3. is $0$ for all $N$

4. is $\infty$ for all $N$

Radius of convergence $r=\lim\limits_{n\to\infty}\frac{p(n)}{p(n+1)}\rightarrow1$ ?

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I've expanded uses of the "$\forall$" symbol into text, because tacked after an expression they look awful. I didn't touch the "$\to$", even though it is not in its place either (it should be "$=$") – Marc van Leeuwen Jul 29 '12 at 9:14

You (probably) know that if $p$ and $q$ are polynomials with the same degree and leading coefficients $p_0$ and $q_{\,0}$, then

$$\lim_{x\to \infty}\frac{p(x)}{q(x)}=\frac{p_0}{q_{\,0}}$$

Your line of thought is good. Since $p(n)$ and $p(n+1)$ have the same degree and the same leading coefficient, say $p_0$, then

$$\lim_{n\to \infty}\frac{p(n)}{p(n+1)}=\frac{p_0}{p_0}=1$$

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You have this. For any nonzero polynomial $p$, you have $$\lim_{n\to\infty} \root{n}\of{|P(n)|} = 1.$$ Now invoke the root test.

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