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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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    @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 $|z|$.2013-01-11

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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 $\lim_{n \to \infty} P(n)/n^N$.

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    +1 to Robert for giving a useful hint instead of typing out a full answer (which would have been just as easy for him to do).2012-12-17