Absolute sequence unbounded within radius of convergence

Question

Let $R$ be the radius of convergence of the complex power series $a_nz^n$ with $0<R<\infty$. Show that if $|z|>R$, then the sequence $|a_nz^n|$ is unbounded.

Trying by contradiction:

So suppose it is bounded: $|a_nz^n|<M$ for some $M$ for all $n$. Then I want to show that this leads to the sequence partial sums of $|a_nz^n|$ being bounded and hence converging, i.e. $|z|<R$, and so proving the contrapositive.

I'm not sure how assuming boundedness helps me proceed because the partial sums of $|a_nz^n|$ up to $N$ are less than $MN$ but this diverges.

Furthermore, would this method work - would I need to consider $|z|=R$ also?

Answer 1

Assume that the sequence is bounded and let $r$ such that $|z|>r>R$ then $$|a_n r^n|=|a_nz^n|\left(\frac{r}{|z|}\right)^n\le M \left(\frac{r}{|z|}\right)^n$$ and the geometirc series $\sum_n\left(\frac{r}{|z|}\right)^n$ is convergent which leads to a contradiction.

Obviously this method doesn't work for $|z|=R$ and notice that in the border of the disc of convergence we may found convergence or divergence of the series.

Answer 2

Hint: Suppose that the sequence $(|a_n z^n|)$ is bounded. Let $R'=\frac{R+|z|}{2}$. If $|w|\le R'$, then $\sum a_n w^n$ converges.

Answer 3

Assume that $a_n z^n$ is bounded by $M \geq 0$ and let $w \in \mathbb{C}$ with $|w| < |z|$. Then $|a_n w^n| = |a_n z^n| \cdot |(w/z)^n| \leq M \cdot |w/z|^n$.

Note that $\sum_{n=1}^{\infty} |w/z|^n < \infty$ as $|w/z| < 1$ (geometric series).

By the comparison test, the series $\sum_{n=1}^{\infty} a_n w^n$ converges (absolutely) for all $|w| < |z|$, which implies that the radius of convergence is at least $|z|$.

For $|z| > R$ we derive a contradiction.