Convergence of Taylor Series (as part of finding the region of Conv for a Laurent series)

Question

I'm given a Laurent series problem and to find the largest region of convergence, I need to fine $R_1$ and $R_2$. The Laurent series is $$z^3 - \frac{z}{3!} + \sum_{k\geqslant2} \frac{(-1)^{k}}{(2k+1)!z^{2k-3}}$$ Using the geometric series, I found that $$|z| > 1 = R_1$$ But for R2, I thought comparing it to a Taylor series whose coefficient for k=1 is greater than 1/3! and for k=3 is greater than or equal to 1. However, I have not found anything that works. Is there any other ideas that would help find R2 such that R2 > R1?

Answer 1

Your series has nothing to do with the geometric series. It follows from the ratio test that it converges for every $z\ne0$. So, $R_1=0$ and $R_2=\infty$.

Answer 2

Using formulae $r=\limsup|a_{-n}|^{1/n}=\limsup1/(2k+1)!^{1/(2k+3)}=1/\infty=0$ and $R=1/\limsup|a_{n}|^{1/n}=1/0=\infty$.