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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

$$R = \sup \{r \ge 0: \text{the sequence } (c_nr^n) \text{ is bounded}\}.$$

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.

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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.

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  • $\begingroup$ Hi! I $think$ I understand, but could I please run my whole reasoning by you to check? My reasoning is this: Let $c_n$ be a sequence and let's say $R=1$. Take $r=0.5$. Then $c_n (0.5)^n$ is bounded. Let $z_0=0$. The theorem tells us that $\sum_{n=0}^{\infty} c_nz^n$ converges absolutely for any $z$ with $|z|<1$; pick $z = 0.5$, Then $\sum_{n=0}^{\infty} c_n (0.5)^n$ converges. But here I made the mistake; the conclusion we can draw from here is that $\sum_{n=0}^{\infty} c_nr^n$ converges for "almost every" $r$ for which $c_nr^n$ is bounded (for any $r$ with $|r|<1$). (continued) $\endgroup$ – Ovi Dec 29 '18 at 14:45
  • $\begingroup$ (continued) But we cannot draw the conclusion that "If $c_n r^n$ is bounded, then $\sum_{n=0}^{\infty} c_n r^n$ converges, because it may be the case that $c_n R^n$ is bounded, but $\sum_{n=0}^{\infty} c_n R^n$ does not converge. $\endgroup$ – Ovi Dec 29 '18 at 14:45
  • $\begingroup$ Yes, that is correct. The sum will converge for every $r$ with modulus strictly less than $R$, but not necessarily on the circle $|r|=R$. $\endgroup$ – Ross Millikan Dec 29 '18 at 15:26
  • $\begingroup$ Thank you! ${}{}{}$ $\endgroup$ – Ovi Dec 29 '18 at 16:06
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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 $r=1$.

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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.

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