1. Nov 20, 2015

### geoffrey159

1. The problem statement, all variables and given/known data
Given the power serie $\sum_{n\ge 0} a_n z^n$, with radius of convergence $R$, if there exists a complex number $z_0$ such that the the serie is semi-convergent at $z_0$, show that $R = |z_0|$.

2. Relevant equations

3. The attempt at a solution

Firstly, since $\sum_{n\ge 0} a_n z_0^n$ is not absolutely convergent, then $R \le |z_0|$.
Secondly, $\sum_{n\ge 0} a_n z_0^n$ is convergent so $(a_n z_0^n)_n$ is bounded as it tends to zero, so $|z_0| \le R$.
So $R = |z_0|$.

Is it correct ?

Last edited: Nov 20, 2015
2. Nov 20, 2015

### andrewkirk

The proof may be missing something.
What is the justification for this statement? Failure to absolutely converge does not entail failure to converge, and radius of convergence delimits the disc on which ordinary convergence happens, not absolute convergence.
I don't recognise the symbol string $(a_n z_0^n)_n$. What do you intend it to mean?

3. Nov 20, 2015

### geoffrey159

The radius of convergence delimits the disc where absolute convergence happens, and absolute convergence implies ordinary convergence.
Therefore, if the serie fails to converge absolutely in $z_0$, which is the case because it is semi-convergent in $z_0$, then $R\le |z_0|$

I meant the sequence with general term $u_n = a_nz_0^n$

4. Nov 21, 2015

### Samy_A

Agreed.
I don't understand why the fact that $(a_n z_0^n)_n$ is bounded implies $|z_0| \le R$.
But the above quote is correct if you remove part of it.

5. Nov 21, 2015

### geoffrey159

My definition of radius of convergence is the supremum, if it exists, of the set $S= \{ r \ge 0 \text{ s.t the sequence } (a_n r^n)_n \text{ is bounded }\}$. Also, for any $r\in S$, $[0,r[$ is in this set.
With that definition, the serie converges absolutely inside the disc of convergence, and outside the disc of convergence, the sequence $(a_n z^n)_n$ is not bounded .
So if $(a_n z_0^n)_n$ is bounded, then by contraposition, $|z_0| \le R$.

6. Nov 21, 2015

### Samy_A

Yes, you are right. I never saw that definition for radius of convergence, but it is equivalent to the one I knew.

7. Nov 21, 2015

### geoffrey159

Thank you for looking at it