How can I find the radius of convergence of the following ?
$$\sum_{n=0}^\infty \frac{n^n}{n!}(z-\pi i)^n $$
The problem is that it involves the complex number $\pi i$, and that is giving me a lot of trouble.
Here is how I have gone this far:
$$\frac{(n+1)^{n+1}}{(n+1)!}\times\frac{n!}{n^n}(z-\pi i)=\bigg(1+\frac{1}{n}\bigg)^n (z-\pi i)$$
Is the radius of convergence then $\frac{1}{e}$?
