Q:

Nice exercise with Rouche theorem

let $P_n(z) = 1 + z/1! + z^2/2! + ... + z^n/n!$
prove that for all $R>0$ sufficiently large, there exists $N\in \mathbb N$ such that for all $n \ge N$, all the roots of $P_n$ lie in $D(0,R)$
I know this has to do with Rouche's theorem. 
since $P_n$ converges uniformly to $e^z = f(z)$ for all $\epsilon > 0$  there exists $N$ such that for $n \ge N$: $\sup |P_n(z) - f(z)| < \epsilon$
I want the bound $|P_n(z) - f(z)| < |f(z)$ somewhere to deduce that $P_n$ has the same number of roots (=zero) as $f$ inside $D(0,R)$
I am not sure how to make an argument and i don't have a clear idea about how to proceed
please help and try to give as much details as possible because I am having trouble doing arguments
thanks 

A:

I think this result, as stated, is false. However, we can prove, using Rouche, a sort of "opposite" of the result you are looking for, namely:

For all $R>0$, there is $N$ s.t. for all $n\geq N$, all the zeros of $P_n$ lie outside of $D(0,R)$. 

Proof:
We know $|f(z)|=|e^{x+iy}|=e^x$. 
Let $n\in \mathbb{N}$ and $z\in \partial{D}(0,R)$. We know 
$$
|f(z)|=|e^{x+iy}|=e^x\geq e^{-R}
$$
Also, note that
$$
|f(z)-P_n(z)|=\left|\sum_{k>n}\frac{z^k}{k!}\right|\leq \sum_{k>n}\frac{|z|^k}{k!}= \sum_{k>n}\frac{R^k}{k!}
$$
and since $\sum_{k=0}^n \frac{R^k}{k!}\to e^R$, the "residual" $\sum_{k>n}\frac{R^k}{k!}$ becomes as small as we want. Hence, there is $N$ s.t. for all $n\geq N$, $\sum_{k>n}\frac{R^k}{k!}<e^{-R}$, which implies, for all $z\in \partial{D}(0,R)$, 
$$
|f(z)-P_n(z)|\leq \sum_{k>n}\frac{R^k}{k!}\leq e^{-R}\leq |f(z)|
$$
and by Rouche we conclude $P_n$ has no zeros in $D(0,R)$ for $n\geq N$.

