0
$\begingroup$

There is the following result (not proven) on my notes: If $\sum_{j=-\infty}^\infty a_j (z-z_0)^j$ is converges at some point, then there exist $r_1,r_2\geq 0$ (which can be $\infty$) so that the series is absolutely convergent for $r_1<|z-z_0|<r_2$ and divergent for $|z-z_0|>r_2$ or $|z-z_0|<r_1$. There also exist $r_1',r_2'$ so that the series converges uniformly for $r_1<r_1'\leq |z-z_0|\leq r_2'<r_2$.

Is this result correct? I think I understand that there be convergence and uniform convergence due to the Cauchy-Hadamard theorem, but I cannot see why there should also be absolute convergence. Does this result imply that convergent Laurent series are also absolutely convergent?

$\endgroup$

1 Answer 1

2
$\begingroup$

Remember that a Laurent series is just the sum of two power series, one in $u = (z - z_0)$ and the other in $w = \dfrac 1{z-z_0}$. Each power series has a radius of convergence, say $r_u$ for the series in $u$ and $r_w$ for the series in $w$. Power series are absolutely convergent everywhere inside their radius of convergence (if you look at the proof of Hadamard's formula for the radius of convergence, you will see that it actually proves absolute convergence).

But $$|u| < r_u \implies|z-z_0| < r_u$$

and $$|w| < r_w \implies \dfrac 1{|z-z_0|} < r_w \implies \dfrac 1{r_w} < |z - z_0|$$

So setting $r_2 = r_u$ and $r_1 = \frac 1{r_w}$ gives that the Laurent series is absolutely convergent on $r_1 \le |z - z_0| \le r_2$.

$\endgroup$
3
  • $\begingroup$ Thank you very much for your answer, but that explains the uniform convergence due to the Cauchy-Hadamard Theorem for power series, as I said, but this power series theorem doesn't assure absolute convergence right? I cannot see why the result in my notes claims absolute convergence, and whether that implies that convergent Laurent series are also absolutely convergent or not. Thanks! $\endgroup$
    – Oski
    Jun 2, 2020 at 10:04
  • $\begingroup$ sorry - I actually wrote this for absolute convergence, but when I finished, I glanced at your question and saw you talking about unitform convergence and thought that I had gotten that wrong, so quickly changed to talk about uniform convergence. Absolute convergence is a little easier, because all power series are absolutely convergence everywhere inside their radius of convergence. It is only on the boundary that absolute convergence can fail. So once again, as the sum of two power series, Laurent series are absolutely convergent everywhere in the interior of the annulus. $\endgroup$ Jun 2, 2020 at 11:03
  • $\begingroup$ Thanks a lot! I guess that the Cauchy-Hadamard theorem was incomplete in my notes then :') $\endgroup$
    – Oski
    Jun 2, 2020 at 11:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .