Find the radius of convergence for \[ f(z) \da \sum_{k\in \ZZ} 2^{-\abs{k}}z^k .\]
Solution
Break this up into a principal part at \(z=0\) and a holomorphic part: \[ f(z) = f_1(z) + f_2(z) \da \sum_{k\geq 1} 2^{-k}z^{-k} + \sum_{k\geq 0} 2^{-k}z^k .\]
Using the ratio test: \[ f_1(z) < \infty &\impliedby \limsup_k \abs{2^{-k}z^{-k}}^{1\over k} < 1 \iff \limsup_k \abs{1\over 2z} < 1 \iff {1\over 2}< \abs{z} \\ f_2(z) < \infty &\impliedby \limsup_k \abs{2^{-k}z^{k}}^{1\over k} < 1 \iff \limsup_k \abs{z\over 2}< 1 \iff \abs{z} < 2 .\]
So \(f\) converges on \({1\over 2}< \abs{z} < 2\).