Using Rouché's theorem to bound zeroes of partial sums of $e^z$ I'm having trouble with a question from an exam I'm studying for:
"
Prove that all roots of the polynomial $\sum\limits_{k=0}^n \frac{z^k}{k!}$ (for $n\geq 1$) are in the annulus $\{z: \frac{n}{e}<\vert z\vert<2n \}$.
Hint: use the inequality $n^ne^{-n}<n!$"
These kind of questions are usually solved by using Rouché's theorem, so I assume that's the case here as well, but I'm having trouble choosing the right functions to use and bounding their norms correctly. I feel like I should be comparing $e^z$ and $\left(e^z - \sum\limits_{k=0}^n \frac{z^k}{k!}\right)$ on the circle with radius $\frac{n}{e}$ since e^z has no roots there, so maybe that will imply the polynomial also has no roots there. And then maybe using some other argument to show that all the polynomial's roots are inside the larger disk of radius $2n$.
 A: Showing that all roots are inside the disk of radius $2n$ is easy - that's a simple application of Rouche's theorem, with the $\frac{z^n}{n!}$ term dominating the sum on the circle.
Showing that no roots are inside the small circle - what you have looks like a good start. How about the Taylor's theorem error estimate to get a size for the difference? That leads to comparing $e^{-|z|}$ and $\frac{e^{|z|}|z|^{n+1}}{(n+1)!}$ (Lagrange form), so we want $|z|^{n+1}e^{2|z|} < (n+1)!$. That's not true at $|z|=\frac{n}{e}$, with the exponentials working against us - we get about $\left(\frac{n}{e}\right)^{n+1}$ from the power of $z$ and $e$ times that from the factorial, so multiplying by the growing exponential breaks it. It's plausible that the result might be true, but this estimate isn't strong enough to prove it.
The thing about this one? The lower bound on the size of the roots is pretty tight. Running it numerically for $n=9$, there's a root at about $-3.33$, and the bound $\frac{9}{e}$ says there are no roots with absolute value less than about $3.31$. We'll need a very strong estimate of some kind to make that inner circle work. I'm not sure what that will be right now - I'll have to think on it some more.
Update: following the links in the comment above leads to the  Szegő curve $|z\exp(1-z)|=1$. For $\frac ne$ to be a lower bound for the size of the roots, the circle $|z|=\frac1e$ would have to lie entirely inside the curve, where $|z\exp(1-z)| < 1$. At $-e^{-1}$, we get $|-e^{-1}\exp(1+e^{-1})|=\exp(e^{-1})>1$. That point is outside the curve. The proposition you were asked to prove is false.
Indeed, for $n=11$, the negative real root is at approximately $-3.90545$, while $\frac{11}{e}\approx 4.04667$.
