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

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 $$|z - z_0|, so $$\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.

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.
• 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) – Ovi Dec 29 '18 at 14:45
• (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. – Ovi Dec 29 '18 at 14:45
• Yes, that is correct. The sum will converge for every $r$ with modulus strictly less than $R$, but not necessarily on the circle $|r|=R$. – Ross Millikan Dec 29 '18 at 15:26
• Thank you! ${}{}{}$ – Ovi Dec 29 '18 at 16:06
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 $$r is not equivalent to $$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.