Radius of convergence of the series $\sum\limits_{-\infty}^{\infty}(2^{-n}+4^{-n}) z^n$

64 Views Asked by At

I'm trying to find for what values of $z\in\mathbb{C}$ the series $$\sum_{n=-\infty}^{\infty}(2^{-n}+4^{-n})z^n$$ converges. My main methods are the nth root test and the ratio test. I believe it can be done using these. However I am unsure due to the negative infinity in the series, I'm not sure how I can apply either of the two tests to this series.

My initial thought was to split this up into two series: $$\sum_{n=-\infty}^{\infty}(2^{-n}+4^{-n})z^n=\sum_{n=0}^{\infty}(2^{-n}+4^{-n})z^n+\sum_{n=1}^{\infty}(2^{n}+4^{n})z^{-n}$$ however I'm not sure if that is useful or indeed allowed. Then using the ratio test I achieve that the radius of convergence for the first sum is $1$, i.e., convergent for $|z|<1$ and divergent for $|z|>1$ and undetermined for $|z|=1$. However I do not know how to go about the other limit with the exponent of the $z$ being $-n$. Can anyone advise on a method which can allow one to use the tests on these type of series?