Proof $\sum_{k=0}^\infty a_k(z-z_0)^k$ diverges for $|z-z_0| > 1/a$, converges for $|z-z_0| < 1/a$ and absolutely converges for $a=0$?

Question

Let $z_0 \in \mathbb{C}$ and $a_k \neq 0$ for all $k \in \mathbb{N}$.

We look at the power series $\sum_{k=0}^\infty a_k(z-z_0)^k$ and let $\lim_{k \to \infty} |\frac{a_{k+1}}{a_k}| = a$ exist.

How can one prove that for $a \neq 0$ the series diverges for $|z-z_0| > 1/a$ and absolutely converges for $|z-z_0| < 1/a$? And that it for $a=0$ for all $z \in \mathbb{C}$ it absolutely converges?

This is what I have, but the problem is that I don't show that it diverges for $> 1/a$ and converges for $< 1/a$. I also don't know to show that it converges for $a= 0$ for all $z \in \mathbb{C}$...

Let the power series converge for $z_1 \neq z_0$ and let $|z-z_0| < |z_1-z_0|$. From the convergence of $\sum_{k=0}^\infty a_k(z_1 - z_0)^k$ it follows that $\lim_{k \to \infty} a_k(z_1-z_0)^k = 0$. Therefore, an $M > 0$ exists, such that for all $k \in \mathbb{N}$ it follows that $|a_k(z_1-z_0)^k| \leq M$.

This is what I have, but the problem is that I don't show that it diverges for $> 1/a$ and converges for $< 1/a$. I also don't know to show that it converges for $a= 0$ for all $z \in \mathbb{C}$

So, we get

$$|a_k(z-z_0)^k| = |a_k(z_1-z_0)^k| |\frac{z-z_0}{z_1-z_0}|^k \leq M |\frac{z-z_0}{z_1-z_0}|^k$$

Because $|\frac{z-z_0}{z_1-z_0}| = q < 1$ the convergence follows via the comparative criterion of $\sum_{k=0}^\infty a_k(z-z_0)^k$

Answer 1

Suppose $|z - z_0| > 1/a$. Since $|a_{k+1}|/|a_k| \to a$ as $k \to \infty$ it follows that

$$\lim_{k \to \infty}\frac{|a_{k+1}(z-z_0)^{k+1}|}{|a_k(z - z_0)^k|} = \lim_{k \to \infty}\frac{|a_{k+1}|}{|a_k|}|z - z_0|= a |z-z_0|> 1,$$

and, given $r$ such that $a|z-z_0| > r > 1$, there exists $N \in \mathbb{N}$ such that for all $k \geqslant N$ we have

$$\frac{|a_{k+1}(z - z_0)^{k+1}|}{|a_k(z-z_0)^k|} > r$$

By induction, it follows that for all $j \geqslant 1 $ we have

$$\frac{|a_{N+j}(z - z_0)^{N+j}|}{|a_N(z-z_0)^N|} > r^j,$$

and for all $n = N+j > N$,

$$\tag{*}|a_n(z-z_0)^n| > r^n \frac{|z- z_0|^N}{r^N}$$

Since $r > 1$ and $|z - z_0| > 1/a > 0$, the RHS of (*) and, consequently, $|a_n(z-z_0)^n|$ diverge to $\infty$ as $n \to \infty$. Hence, the power series diverges by the term test.

On the other hand, if $|z-z_0| < 1/a$ then repeating the argument (with inequalities reversed) there exists $r < 1$ such that for all $n > N$,

$$|a_n(z-z_0)^n| < r^n \frac{|z- z_0|^N}{r^N},$$

and the power series must converge by comparison with a convergent geometric series.