Solution

${\displaystyle \sec \pi z={\frac {1}{\cos \pi z}}}$. By using the definition ${\displaystyle \cos z={\frac {e^{iz}+e^{-iz}}{2}}}$, we get that ${\displaystyle \cos(z)=0}$ if and only if ${\displaystyle e^{-y}e^{ix}=-e^{y}e^{-ix}}$. It is not hard to show that this happens if and only if ${\displaystyle y=0}$ and ${\displaystyle x={\frac {(2n+1)\pi }{2}}}$. Therefore, the only zeros of ${\displaystyle \cos \pi z}$ all occur on the real axis at integer distances away from 1/2. Therefore, ${\displaystyle \sec \pi z}$ is analytic everywhere except at these points.

Our Taylor series ${\displaystyle \sum _{n=0}^{\infty }a_{n}(z+i)^{n}}$ is centered at ${\displaystyle z=-i}$. By simple geometry, the shortest distance from ${\displaystyle z=-i}$ to ${\displaystyle z=1/2}$ or ${\displaystyle z=-1/2}$ (the closest poles of ${\displaystyle \sec \pi z}$) is ${\displaystyle R={\sqrt {1/2^{2}+1^{2}}}={\frac {\sqrt {5}}{2}}}$. This is the radius of convergence of the Taylor series.

From calculus (root test), we know that ${\displaystyle \lim \sup _{n\to \infty }|a_{n}|^{1/n}=1/R}$. Therefore, ${\displaystyle \lim \sup _{n\to \infty }|a_{n}|^{1/n}={\frac {2}{\sqrt {5}}}}$.