# Laurent series region of convergence

I'm trying to find region of convegence of the Laurent series $$\sum_{n=-\infty}^\infty \frac{z^n}{2^{-n}+4^n}$$

I expand it to $$\sum_{n=0}^\infty \frac{z^n}{2^{-n}+4^n} + \sum_{n=1}^{\infty} \frac{z^{-n}}{2^{n}+4^{-n}}$$

After I use ratio test to get ROC for the first term (analogously for the second but the problem is in limits) $$\lim_{n\to\infty} \bigg(\frac{z^{n+1}}{2^{-n-1}+4^{n+1}}\times \frac{2^{-n}+4^{n}}{z^{n}}\bigg) = \lim_{n\to\infty} \frac{2^{-n}+4^{n}}{2^{-n-1}+4^{n+1}}z$$ and the problem for me here is to find the limit. If I understand how the limit is done for the first term, I will apply it for the second

• will it be equal to 1/4? – Lautern Feb 25 '18 at 12:20

$$\lim_{n\to\infty} \frac{2^{-n}+4^{n}}{2^{-n-1}+4^{n+1}}|z| =\lim_{n\to\infty} \frac{2^{-3n}+1}{2^{-3n-1}+4}|z| =\frac{|z|}4$$ etc. So you need $|z|<4$ for convergence. You need to look at the negative $n$ part of the series for the inner radius of the annulus.