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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?

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

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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.

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  • $\begingroup$ But you should include absolute value bars around the terms, because they are complex, right? And then you get that the absolute value of a subsequence converges to a number greater than one, hence the sequence of the absolute value of the terms doesn't converge to zero, and my proof below is the rest of the work. $\endgroup$ – Steven Wagter Jan 12 at 8:32
  • $\begingroup$ Yes I forgot those. I just wanted to show that you can use that limsup is the limit of a subsequence. $\endgroup$ – Math_QED Jan 12 at 23:54
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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.

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    $\begingroup$ This shows that the root test is actually rather poor in practice: Most series we encounter in practice and for which convergence is not evident have terms converging to 0, so if the series diverges, we will not learn it from the test. $\endgroup$ – Andrés E. Caicedo Jan 11 at 20:42
  • $\begingroup$ @AndrésE.Caicedo thanks for that comment, so you could go as far as to say that the root and ratio tests are rather useless for divergence; you could just check whether the terms don't tend to zero instead? Unless the tests are computationally more convenient of course, which is probably seldom. $\endgroup$ – Steven Wagter Jan 11 at 21:20
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    $\begingroup$ In essence you are right. It would be very useful to have a widely applicable convergence test for which the reason for divergence actually tells us something interesting about the terms of the series. Similarly, the ration and the root tests tell us that a series converges when it can be appropriately compared to a convergent geometric series. The tests give us absolute convergence in that case. Many series we find in practice cannot be compared to geometric series an instead may converge very slowly. Also, we have no good tests for conditional (rather than absolute) convergence. $\endgroup$ – Andrés E. Caicedo Jan 11 at 21:25

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