Does n power of e grow much more faster than its Maclaurin polynomial?

Question

I wonder how to calculate the following limit: $$ \lim_{n\rightarrow\infty}\frac{1+n+\frac{{}n^{2}}{2!}+\cdots +\frac{n^{n}}{n!}}{e^{n}} $$ In the first sight, I think it should be zero, because exponential function is much faster than polynomial. But the upper of the expression is the Maclaurin polynomial of $e^{n}$. With the growth of n, it approaches to $e^{n}$. Consider there is a roughly way to estimate the remainder of $e^{n}$ rather than $$ R_{n+1}(n)=\frac{\xi^{n+1}}{(n+1)!} \ for \ \ \xi\in(n,+\infty) $$ Because $$ \frac{1+n+\frac{{}n^{2}}{2!}+\cdots +\frac{n^{n}}{n!}}{e^{n}}=1-\frac{R_{n+1}(n)}{e^{n}} $$ But it's hard to continue.

Answer 1

Let a be the difference error between $e^n$ and maclaurins series the value of the expression now becomes $1-\frac a{e^n}$ and now as n tends to $\infty $ the difference error a tends to zero and $e^n$ tends to $\infty$ and therefore the limit tends to 1 which implies that they both grow at same rate.