estimation - a limit about exponential function

$\lim_{n\rightarrow\infty}\frac{1+\frac{n}{1!}+\cdot+\frac{n^n}{n!}}{e^n}=\frac12$

Taking the first $n$ terms of the Taylor series of $e^n$ as the numerator, the limit is true or false? How to prove?

Answer

Assuming that we work

$$a_n=e^{-n}\sum_{k=0}^n \frac{n^k}{k!}$$ by the definition of the incomplete gamma function
$$a_n=\frac{\Gamma (n+1,n)}{n \Gamma (n)}$$
We have the relation $$\Gamma (n+1,n)=n \,\Gamma (n,n)+e^{-n}\, n^n$$ which makes

$$a_n=\frac{ n^{n-1}}{e^n\,\Gamma (n)}+\frac{\Gamma (n,n)}{\Gamma (n)}$$ The first term tends to $0$ when $n$ becomes large; to prove it, take its logarithm and use Stirling approximation to get
$$\log\left(\frac{ n^{n-1}}{e^n\,\Gamma (n)} \right)=-\frac{1}{2} \log \left({2 \pi n}\right)-\frac{1}{12 n}+O\left(\frac{1}{n^{5/2}}\right)$$

For the second term, if you look here, you will notice the asymptotics

$$\Gamma(n,n) \sim n^n e^{-n} \sqrt{\frac{\pi}{2 n}}$$ So, neglecting the first term, we have, for large $n$
$$a_n\sim \frac{ n^n e^{-n} }{\Gamma(n)}\sqrt{\frac{\pi}{2 n}}$$ Take logarithms and use Stirling approximation to get
$$\log(a_n)=-\log (2)-\frac{1}{12 n}+O\left(\frac{1}{n^{5/2}}\right)$$ Continue with Taylor
$$a_n=e^{\log(a_n)}=\frac{1}{2}-\frac{1}{24 n}+O\left(\frac{1}{n^{2}}\right)$$
If you use the better asymptotics given in the link $$\Gamma(n,n) = n^n e^{-n} \left [ \sqrt{\frac{\pi}{2 n}} - \frac{1}{3 n} + O\left ( \frac{1}{n^{3/2}} \right ) \right ]$$ doing the same, you should end with

$$a_n=\frac 12-\frac{1}{3 \sqrt{2 \pi n} }-\frac{1}{24 n}+O\left(\frac{1}{n^{3/2}}\right)$$