prove that all the roots of $\sum_{k=0}^{n}\frac{z^{k}}{k!}$ are in $\{z:\frac{n}{e}<|z|<2n\}$. I thought about using Rouché's theorem but there is no function which I can compare to. Does anyone have an idea?
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Perhaps you could show that all $n$ roots are in the outer disk and none are inside :-). – copper.hat Aug 04 '20 at 17:00
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for example $\frac{a^n}{n!} \ge \frac{a^k}{k!}, a \ge n \ge k$ so writing $2^n=1+(2^{n-1}+....2+1)$ one imediately gets that if $b \ge 2n$ and $b=2a, a \ge n$ the leading term $(2a)^n/n=2^n a^n/n!=(1+(2^{n-1}+....2+1))a^n/n!$ strictly dominates the other terms for $|z|=b \ge 2n$ so $z$ is not a root when $|z| \ge 2n$; for the other inequality show that now $1$ dominates the expression for $|z| \le n/e$ – Conrad Aug 04 '20 at 17:21