1
$\begingroup$

Show that the radius of convergence of a power series $\sum_{n=0}^{\infty}a_nz^n$ is given by the supremum of the set of real numbers $r\geq 0$ with the property that there exists a bound M such that $|a_n|r^n \leq M$ for all $n$ (M may depend on r)

I am not sure how to do this proof. Any help/hint is appreciated. Thank you.

$\endgroup$
1

1 Answer 1

1
$\begingroup$

Let $$ r_0 = \sup \{\; r \ge 0 : \mbox{ there exists } M \ge 0 \mbox{ such that } |a_n|r^n \le M \mbox{ for all } n \ge 0 \} $$ I'll leave the special cases where $r_0=0$ or $r_0=\infty$ to you. I'll assume $0 < r_0 < \infty$.

  • If $|z| < r_0$, then $\sum_{n}a_n z^n$ converges absolutely by the definition of $r_0$. Hence, the series converges absolutely for any fixed $z$ for which $|z| < r_0$.

  • If $|z| > r_0$, then $\{ |a_n||z|^n \}$ is an unbounded sequence, which means that the general term of the series $\sum_{n}a_n z^n$ does not converge to $0$ and, thus, cannot converge conditionally or absolutely.

By the definition of the radius of convergence, $r_0$ is the radius of convergence.

$\endgroup$
2
  • $\begingroup$ Don't quite see your first claim. Why does it converge absolutely? $\endgroup$
    – user223391
    Commented Jan 19, 2017 at 23:28
  • 1
    $\begingroup$ @ZacharySelk : Let $|z| < r_0$. Then, for any $r$ such that $|z| < r < r_0$, it must be true that $|a_n|r^n \le M$. That means $|a_n||z|^n = |a_n|r^n\left(\frac{|z|}{r}\right)^n \le M\rho^n$ where $\rho < 1$, which makes $\sum a_n z^n$ absolutely convergent. $\endgroup$ Commented Jan 19, 2017 at 23:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .