Consider the polynomial of degree $n$, we have $P_n(z)=z+\frac{z^2}{2}+...+\frac{z^n}{n!}$.
Arrange all of its roots $(z_{n,1},z_{n,2},...,z_{n,n})$ in increasing order of magnitude $0=|z_{n,1}|\leq |z_{n,2}|\leq ... \leq |z_{n,n}|.$
Show that $\lim\limits _{n\to\infty}|z_{n,2}|=2\pi$.
Thoughts so far:
Very generally, I would like to show that $|z_{n,2}|$ < $|z_{n,3}|$ so that we can have r>0 so that the disc |z|< r containing only 2 roots of $P_{n}$: $z_{n,2}$ and 0. Then I would like to integrate $\frac{1}{P_{n}}$ over |z| = r, and use the residue theorem to get an expression in terms of $z_{n,2}$, and then solve for $z_{n,2}$. Then, I'd take the limit as n $\rightarrow \inft However, I'm not having a lot of success with this method... any suggestions for another way to approach the problem? Thanks!