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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

$$R = sup r ge 0: textthe sequence (c_nr^n) text is bounded.$$

Then the power series $sum_n=0^inftyc_n(z-z_0)^n$ 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 $sum_n=0^inftyc_n(z-z_0)^n$ converges absolutely whenever $|z - z_0|<R$, so $sum_n=0^inftyc_nr^n$ converges whenever $r < R$, so $sum_n=0^inftyc_nr^n$ converges whenever $c_n r^n$ is bounded.

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.

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 $r=1$.

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.