# The radius of convergence of the power series $\sum_{0}^{\infty}P(n)x^n$

I came across the following problem:

Let $P(x)$ be a non-zero polynomial of degree $N.$ The radius of convergence of the power series $\sum_{0}^{\infty}P(n)x^n$
(a)depends on $N,$
(b)is $1$ for all $N,$
(c)is $0$ for all $N,$
(d)is $\infty$ for all $N.$

My attempts:
I see that $lim_{n\to\infty}P(n+1)/P(n)=1$,because if I take $N=2$,then we can take $P(n)=a_0+a_1n+a_2n^2$ and $P(n+1)=a_0+a_1(n+1)+a_2(n+1)^2$ and hence $\frac {P(n+1)}{P(n)}=1+\frac {a_1+(2n+1)a_2}{P(n)}$ which tends to $1$ as $n$ tends to $\infty$. From here,Can i say that (b)is the right choice? Am i going in the right direction? Please comment.

Can someone point me in the right direction? Thanks in advance for your time.

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No, $\lim_{n \to \infty} P(n+1)/P(n)$ is not $\infty$. Try again. – Robert Israel Dec 17 '12 at 6:33
Intuitively, as $n \to \infty$ we have $|P(n)| \to \infty$ polynomially if $N \geq 1$, and if $x>0$ then $x^n \to 0$ geometrically if $x<1$ (so the series converges for $x<1$) and $x^n \to \infty$ geometrically if $x>1$ (so the series diverges for $x>1$). The radius of convergence must be exactly $1$ since geometric convergence or divergence dominates polynomial divergence. – Antonio Vargas Jan 10 '13 at 18:28
@Antonio: It is sufficient for power series. – Jonas Meyer Jan 11 '13 at 3:52
@AntonioVargas: Actually, now I should backtrack a bit. The possible exception is on the boundary of convergence, but that doesn't affect using your intuition because you were working with all $x<1$ (and it is already clear in this case what happens when $|x|=1$.) In general, one can say that if $a_nz^n\to 0$ for all $|z|<C$, then the radius of convergence is at least $C$. – Jonas Meyer Jan 11 '13 at 4:59
@JonasMeyer right, it seems that either $\limsup |a_n z^n| = \infty$ or $= 0$ depending on whether $|z|$ is bigger or smaller than some value; at least you will be able to determine where the boundary is. In fact if $a_n z^n = O(1)$ for some $z$, then the radius of convergence is at least $|z|$. – Antonio Vargas Jan 11 '13 at 5:14

Hint: What is $\lim_{n \to \infty} P(n+1)/P(n)$? You might want to try induction on the degree, using l'Hopital's Rule. Or you could consider $\lim_{n \to \infty} P(n)/n^N$.
sir, i see that $lim_{n\to\infty}P(n+1)/P(n)=\infty.$ – user52976 Dec 17 '12 at 5:57