# Laurent Series absolute, almost uniform convergence

Suppose that $$\sum_{n=-\infty}^{\infty} c_n z^n$$ converges absolutely when $z=z_0$, where $z_0$ is a nonzero complex number. How can one show that there is an $R \geq |z_0|$ such that $$\sum_{n \geq 0} c_n z^n$$ converges absolutely and almost uniformly on the disk $|z|\leq R$? Similarly, how can we show there is a nonzero real number $r\leq |z_0|$ such that $$\sum_{n < 0} c_n z^n$$ converges absolutely and almost uniformly on $|z| \leq r$?

What I've tried thus far (for the former question; I figure the second will be similar):

The absolute convergence of $$\sum_{n=-\infty}^{\infty} c_n z_0^n$$ implies the absolute convergence of $$\sum_{n \geq 0} c_n z_0^n,$$ which implies standard convergence, so that $$\lim_{n->\infty} c_n z_0^n = 0.$$ Thus we can choose sufficiently large $N$ so that $|c_n z_0 ^n | \leq 1$ for all $n > N$, and so $|c_n| \leq \frac{1}{|z_0|^n}$. So, $$\sum_{N+1}^{\infty} |c_n z^n| = \sum_{N+1}^{\infty} |c_n| |z|^n \leq \sum_{N+1}^{\infty} \frac{|z|^n}{|z_0|^n}.$$ The final series converges for $|z| < |z_0|$. Therefore, by the comparison test, the first series converges, i.e., the given series is absolutely convergent.

Now let $M_n = \frac{|z_0|^n}{R^n}$, then $\sum_{n \geq 0} M_n$ converges if and only if $R > |z_0|$. As before, deduce $|c_n z^n| < M_n$ for $|z| \leq R$ and sufficiently large $n$ so that, by Weierstrass M test, $\sum_{n\geq 0}c_n z^n$ is uniformly convergent.

I'm not sure I applied Weierstrass correctly, and I'm not sure I successfully showed absolutely convergence and "almost" uniform convergence on the entirety of the disk $|z|\leq R$. (In case it's not clear I'm also not sure what's meant by almost uniform convergence.)

• Are you familiar with the radius of convergence of a series $a_0,a_1,....$ (root test)? – copper.hat Oct 15 '16 at 15:16
• I am, should I be applying it to $\frac{|z|^n}{|z_0|^n}$? One issue I'm having in particular is that most tests will fail for the case $|z|=R$ (i.e. fail on the disk boundary), which I believe would also happen with root test, no? – parssignum Oct 15 '16 at 16:08
• Well, if you define $R$ using the $\limsup$ way, then the sequence converges if $|z| < R$ and diverges if $|z| > R$. Since you know the sequence converges at $z_0$, you know that $|z_0| \not > R$, that is $|z_0| \le R$. (Just to be sure, I am defining ${1 \over R} = \limsup_n \sqrt[n]{|c_n|}$.) – copper.hat Oct 15 '16 at 16:14