I wish to prove that the absolute values of the roots of the polynomial $p(x)=1+x+x^{2}/2!+...+x^{n}/n!$ are between $n/2e$ and $2n$. I was able to prove the upper bound using the following:
Claim: If $a_{n}x^{n}+\dots+a_{1}x+a_0$ is a polynomial, then its roots are bounded from above by $2\max\left\{\left(\frac{|a_i|}{|a_n|}\right)^{1/(n-i)}:\;0\leq i<n\right\}$.
It works perfectly, because $\frac{n!}{i!}<n^{n-i}$, as one can see immediately by expanding the left hand side.
To achieve the lower bound, I tried the following idea - if $x$ is a root of $p(x)$, then $1/x$ is a root of $x^{n}p(1/x)$. Thus I need to bound the roots of $q(x)=x^{n}+x^{n-1}+\dots+x/(n-1)!+1/n!$ from above by $2e/n$. I tried to use the same claim, but now things are different, because now $\left(\frac{|a_i|}{|a_n|}\right)^{1/(n-i)}=\left(\frac{1}{(n-i)!}\right)^{\frac{1}{n-i}}$, which equals $1$ when $i=n-1$, which is certainly not smaller than $e/n$.
Am I missing something? I really can't think of any other idea, or in other words, I'm not sure how I would use Rouche's theorem here, so any hints or references towards this lower bound would be very useful, thank you very much!
