Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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

2 Answers 2

up vote 3 down vote accepted

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

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

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.