Why does the criterion for convergence of a power series not imply every series with bounded terms converges?

Question

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.

Answer 1

The problem comes in the last step. Just because $\sum_{n=1}^\infty c_nr^n$ converges with $r \lt R$ you cannot conclude that $\sum_{n=1}^\infty c_nR^n$ converges. As an example, let $c_n=1$ for all $n$. We note that $R=1$ here. $c_nR^n=1$, so is bounded. For any $r \lt 1$, $\sum_{n=1}^\infty c_nr^n$ converges absolutely, but $\sum_{n=1}^\infty c_nR^n$ does not converge.

Answer 2

The condition $r<R$ is not equivalent to $c_nr^n$ being bounded. It is possible that $c_nr^n$ is bounded for $r=R$ as well, and in that case we cannot conclude that the series converges.

Answer 3

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 $r=1$.