# Radius of convergence of $\sum\limits_{-\infty}^{\infty}3^{-|n|}z^{2n}, z \in \mathcal{C}$

I want to find out the radius of the following power series of a complex variable:

$$\displaystyle \sum_{-\infty}^{\infty} 3^{-|n|} z^{2n}, z \in \mathbb{C}$$

The ration test $$\displaystyle \lim_{n \to \infty}\left|\frac{a_m}{a_{m+1}}\right|$$ gives me the radius of convergence as 3 . But I am not really sure how to handle the negative power terms. Should I group same co-efficient terms, and complete the square? But then, I don't get a polynomial in a single variable.

Also, how do I evaluate the sum?

• As far as I know, the radius of convergence is only defined for series with positive indices, not series with negative exponents. en.wikipedia.org/wiki/Radius_of_convergence Mar 31, 2013 at 7:01

Your series is $$1 + \sum_{n=1}^\infty \bigg(\frac{z^2}3\bigg)^n + \sum_{n=1}^\infty \bigg(\frac1{3z^2}\bigg)^n,$$ which converges when both $$|z^2/3|$$ and $$|1/3z^2|$$ are less than $$1$$, that is, when $$1/\sqrt3<|z|<\sqrt3$$.