I am reading Complex Made Simple by David C. Ullrich. There is a result from which I am deducing bogus conclusions, so I must be misunderstanding it somehow:
Lemma 1.0. Suppose $(c_n)_{n = 0}^{\infty}$ is a sequence of complex numbers, and define $R \in [0, \infty]$ by
$$R = \sup \{r \ge 0: \text{the sequence } (c_nr^n) \text{ is bounded}\}.$$
Then the power series $\sum_{n=0}^{\infty}c_n(z-z_0)^n$ converges absolutely and uniformly on every compact subset of the disk $D(z_0, R)$ and diverges at every point $z$ with $|z-z_0|>R$.
My bogus conclusion:
Let $c_n$ be a sequence of complex numbers and suppose that $c_n r^n$ is bounded. Then $\sum_{n=0}^{\infty} c_n r^n$ converges.
My reasoning:
Let $c_n$ be any sequence of complex numbers. The series $\sum_{n=0}^{\infty}c_n(z-z_0)^n$ converges absolutely whenever $|z – z_0|<R$, so $\sum_{n=0}^{\infty}c_nr^n$ converges whenever $r < R$, so $\sum_{n=0}^{\infty}c_nr^n$ converges whenever $c_n r^n$ is bounded.
Best Answer
The fact that $c_n r^n$ is bounded, means that $r$ is in the set we're taking the sup of, so $r \le R$. But the convergence of $\sum_{n=0}^\infty c_n s^n$ is only guaranteed for $s < R$. It could very well be that $r=R$; a simple example is $c_n = (-1)^n$ and $r=1$.