Proof of divergence of power series in Theorem 3.39, baby Rudin

Question

This is theorem 3.39 from baby Rudin. Here {$c_n$} is a complex sequence, $z$ is a complex number.

Given the power series $\Sigma c_nz^n$, put $$\alpha = \text{lim sup} \sqrt[n] {|c_n|}, R = \frac{1}{\alpha}.$$(If $\alpha = 0, R = +\infty;$ if $\alpha =+\infty, R = 0.$) Then $\Sigma c_nz^n$ converges if $|z| < R$, and diverges if $|z| > R$.

Proof Put $a_n = c_nz^n$, and apply the root test:

$$\text{lim sup} \sqrt[n] {|a_n|} = |z|\text{lim sup} \sqrt[n] {|c_n|} = \frac{|z|}{R}.$$

Using notation above, the root test (Theorem 3.33) tests for absolute convergence of the series $\sum|a_n|$. Therefore, if $\text{lim sup} \sqrt[n] {|a_n|} < 1$, the series $\sum|a_n|$ converges and hence $\sum a_n$ (this is true for any complex series; see https://math.stackexchange.com/a/1655838/606584). But why is it true that if $\text{lim sup} \sqrt[n] {|a_n|} > 1$ (or equivalently $|z| > R$), $\sum a_n$ diverges? Since, for as far as I can tell, $\text{lim sup} \sqrt[n] {|a_n|} > 1$ only implies that $\sum|a_n|$ diverges, which doesn't in general imply that $\sum a_n$ (think alternating harmonic series).

So my question is: how is it so that when $|z| > R$, the power series diverges?

Answer 1

Because the terms don't tend to $0$.

Answer 2

Thanks to Jamie's hint, I understand;

Let $a_n = x_ni+y_n$. If $\text{lim}_{n \to \infty} |x_n| = 0$ and $\text{lim}_{n \to \infty} |y_n| = 0$, then $\text{lim}_{n \to \infty} |a_n| = 0$.

Since $\text{lim}_{n \to \infty} |a_n| \neq 0$, either the sequence of the real parts or the sequence of the complex parts of the terms doesn't converge to zero, so either the real or imaginary part of the series diverges, hence the complex series diverges.

Answer 3

Writing $L = \limsup_{n \to \infty} a_n^{1/n}$, you can also note that there is a subsequence $b_n$ such that $1<L = \lim b_n^{1/n}$, and thus $1 < b_n$ for all $n \geq 1$. This subsequence cannot converge to $0$, hence $a_n$ can't either.