Proving Radius of Convergence |z_0| = R for Power Series

[geoffrey159]

Homework Statement

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 $R = |z_0|$.

Homework Equations

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 ?

 

[andrewkirk]

The proof may be missing something.

Firstly, since $\sum_{n\ge 0} a_n z_0^n$ is not absolutely convergent, then $R \le |z_0|$.

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.

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

I don't recognise the symbol string $(a_n z_0^n)_n$. What do you intend it to mean?

 

[geoffrey159]

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.

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 don't recognise the symbol string $(a_n z_0^n)_n$. What do you intend it to mean?

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

 

[Samy_A]

Firstly, since $\sum_{n\ge 0} a_n z_0^n$ is not absolutely convergent, then $R \le |z_0|$.

Agreed.

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 $|z_0| \le R$.

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.

 

[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, $|z_0| \le R$.

 

[Samy_A]

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, $|z_0| \le R$.

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

 

[geoffrey159]

Thank you for looking at it