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?
Best Answer
Because the terms don't tend to $0$.