The radius of convergence of $\sum_{k \geq 0}c_k(z-z_0)^k$

Question

(a) Suppose that the sequence $\{c_k\}$ is bounded. Show that the radius of convergence of $\sum_{k \geq 0}c_k(z-z_0)^k$ is at least 1.

(b) Suppose that the sequence $\{c_k\}$ does not converge to 0. Show that the radius of convergence of $\sum_{k \geq 0}c_k(z-z_0)^k$ is at most 1.

I have already finished the part (a), however, I am confused by part $(b)$. Does this mean $\{c_k\}$ converge to some nonzero numbers or does this mean $\{c_k\}$ does not converge?

And does (a) and (b) implies that if $\{c_k\}$ is bounded and does not converge to zero then the radius is 1?

Any help with how to show the part (b)? Thanks~

Answer 1

(a) If $\{c_k\}$ is bounded, then $\sum_{k=0}^{\infty}c_k(z-z_0)^k$ converges absolutely for $|z-z_0| = r$ for any $0 \le r < 1$. So the radius of convergence is at least $1$.

(b) If $\sum_n a_n$ is a convergent series, then the general term must converge to $0$. So, if $c_k$ does not converge to $0$, then $\sum_{k=0}^{\infty}c_k(z-z_0)^k$ cannot converge to $0$ if $|z-z_0|=1$, from which it follows that the radius of convergence is strictly less than $1$.

Answer 2

It means exactly "does not converge to zero," which includes both "converges to a nonzero number" as well as "does not converge."

In particular, there exists some small $\epsilon > 0$ such that $|c_k| > \epsilon$ for infinitely many $k$.

Then you have $|c_k (z - z_0)^k| > \epsilon |z - z_0|^k$ for infinitely many $k$. If $|z - z_0| > 1$ you then have a divergent series by the root test.

And yes, you can combine (a) and (b) to obtain your last claim.