Evaluating $\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$ I'm supposed to calculate:
$$\lim_{n\to\infty} e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!}$$
By using WolframAlpha, I might guess that the limit is $\frac{1}{2}$, which is a pretty interesting and nice result. I wonder in which ways we may approach it.
 A: I thought that it might be of instructive to post a solution to a generalization of the OP's question.  Namely, evaluate the limit
$$\lim_{n\to\infty}e^{-n}\sum_{k=0}^{N(n)}\frac{n^k}{k!}$$
where $N(n)=\lfloor Cn\rfloor$, where $C>0$ is an arbitrary constant.  To that end we now proceed.

Let $N(n)=\lfloor Cn\rfloor$, where $C>0$ is an arbitrary constant.  We denote $S(n)$ the sum of interest
$$S(n)=e^{-n}\sum_{k=0}^{N}\frac{n^k}{k!}$$
Applying the analogous methodology presented by @SangchulLee, it is straightforward to show that
$$S(n)=1-\frac{(N/e)^{N}\sqrt{N}}{N!}\int_{(N-n)/\sqrt{N}}^{\sqrt{N}}e^{\sqrt{N}x}\left(1-\frac{x}{\sqrt N}\right)^N\,dx\tag7$$
We note that the integrand is positive and bounded above by $e^{-x^2/2}$.  Therefore, we can apply the Dominated Convergence Theorem along with Stirling's Formula to evaluate the limit as $n\to\infty$.
There are three cases to examine.
Case $1$: $C>1$
If $C>1$, then both the lower and upper limits of integration on the integral in $(7)$ approach $\infty$ as $n\to \infty$.  Therefore, we find
$$\lim_{n\to \infty}e^{-n}\sum_{k=0}^{\lfloor Cn\rfloor}\frac{n^k}{k!}=1$$
Case $2$: $C=1$
If $C=1$, then the lower limit is $0$ while the upper limit approaches $\infty$ and we find
$$\lim_{n\to \infty}e^{-n}\sum_{k=0}^{n}\frac{n^k}{k!}=\frac12$$
Case $3$: $C<1$
If $C<1$, then the lower limit is approaches $-\infty$ while the upper limit approaches $\infty$ and we find
$$\lim_{n\to \infty}e^{-n}\sum_{k=0}^{n}\frac{n^k}{k!}=0$$

To summarize we have found that
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}e^{-n}\sum_{k=0}^{\lfloor Cn\rfloor}\frac{n^k}{k!}=\begin{cases}1&,C>1\\\\\frac12&, C=1\\\\0&, C<1\end{cases}}$$
A: Integration by parts yields
$$
\frac{1}{k!}\int_x^\infty e^{-t}\,t^k\,\mathrm{d}t=\frac{1}{k!}x^ke^{-x}+\frac{1}{(k-1)!}\int_x^\infty e^{-t}\,t^{k-1}\,\mathrm{d}t\tag{1}
$$
Iterating $(1)$ gives
$$
\frac{1}{n!}\int_x^\infty e^{-t}\,t^n\,\mathrm{d}t=e^{-x}\sum_{k=0}^n\frac{x^k}{k!}\tag{2}
$$
Thus, we get
$$
e^{-n}\sum_{k=0}^n\frac{n^k}{k!}=\frac{1}{n!}\int_n^\infty e^{-t}\,t^n\,\mathrm{d}t\tag{3}
$$
Now, I will reproduce part of the argument I give here, which develops a full asymptotic expansion. Additionally, I include some error estimates that were previously missing.
$$
\begin{align}
\int_n^\infty e^{-t}\,t^n\,\mathrm{d}t
&=n^{n+1}e^{-n}\int_0^\infty e^{-ns}\,(s+1)^n\,\mathrm{d}s\\
&=n^{n+1}e^{-n}\int_0^\infty e^{-n(s-\log(1+s)}\,\mathrm{d}s\\
&=n^{n+1}e^{-n}\int_0^\infty e^{-nu^2/2}\,s'\,\mathrm{d}u\tag{4}
\end{align}
$$
where $t=n(s+1)$ and $u^2/2=s-\log(1+s)$.
Note that $\frac{ss'}{1+s}=u$; thus, when $s\ge1$, $s'\le2u$. This leads to the bound
$$
\begin{align}
\int_{s\ge1} e^{-nu^2/2}\,s'\,\mathrm{d}u
&\le\int_{3/4}^\infty e^{-nu^2/2}\,2u\,\mathrm{d}u\\
&=\frac2ne^{-\frac98n}\tag{5}
\end{align}
$$
$(5)$ also show that
$$
\int_{s\ge1}e^{-nu^2/2}\,\mathrm{d}u\le\frac2ne^{-\frac98n}\tag{6}
$$
For $|s|<1$, we get
$$
u^2/2=s-\log(1+s)=s^2/2-s^3/3+s^4/4-\dots\tag{7}
$$
We can invert the series to get $s'=1+\frac23u+O(u^2)$. Therefore,
$$
\begin{align}
\int_0^\infty e^{-nu^2/2}\,s'\,\mathrm{d}u
&=\int_{s\in[0,1]} e^{-nu^2/2}\,s'\,\mathrm{d}u+\color{red}{\int_{s>1} e^{-nu^2/2}\,s'\,\mathrm{d}u}\\
&=\int_0^\infty\left(1+\frac23u\right)e^{-nu^2/2}\,\mathrm{d}u-\color{darkorange}{\int_{s>1}\left(1+\frac23u\right)e^{-nu^2/2}\,\mathrm{d}u}\\
&+\int_0^\infty e^{-nu^2/2}\,O(u^2)\,\mathrm{d}u-\color{darkorange}{\int_{s>1} e^{-nu^2/2}\,O(u^2)\,\mathrm{d}u}\\
&+\color{red}{\int_{s>1} e^{-nu^2/2}\,s'\,\mathrm{d}u}\\
&=\sqrt{\frac{\pi}{2n}}+\frac2{3n}+O\left(n^{-3/2}\right)\tag{8}
\end{align}
$$
The red and orange integrals decrease exponentially by $(5)$ and $(6)$.
Plugging $(8)$ into $(4)$ yields
$$
\int_n^\infty e^{-t}\,t^n\,\mathrm{d}t=\left(\sqrt{\frac{\pi n}{2}}+\frac23\right)\,n^ne^{-n}+O(n^{n-1/2}e^{-n})\tag{9}
$$
The argument above can be used to prove Stirling's approximation, which says that
$$
n!=\sqrt{2\pi n}\,n^ne^{-n}+O(n^{n-1/2}e^{-n})\tag{10}
$$
Combining $(9)$ and $(10)$ yields
$$
\begin{align}
e^{-n}\sum_{k=0}^n\frac{n^k}{k!}
&=\frac{1}{n!}\int_n^\infty e^{-t}\,t^n\,\mathrm{d}t\\
&=\frac12+\frac{2/3}{\sqrt{2\pi n}}+O(n^{-1})\tag{11}
\end{align}
$$
A: On this page there is a nice collection of evidence.
I add another proof which also uses the Stirling formula.   
$\displaystyle e^{-n}\sum\limits_{k=0}^n\frac{n^k}{k!} = e^{-n}\sum\limits_{k=0}^n\frac{k^k (n-k)^{n-k}}{k!(n-k)!} \hspace{4cm}$ e.g. here
$\displaystyle \lim\limits_{n\to\infty} e^{-n}\sum\limits_{k=1}^{n-1}\frac{e^k e^{n-k}}{\sqrt{2\pi k (1+\mathcal{O}(1/k))}\sqrt{2\pi (n-k)(1+\mathcal{O}(1/(n-k)))}} $
$\displaystyle = \lim\limits_{n\to\infty} \frac{1}{2\pi}\frac{1}{n}\sum\limits_{k=1}^{n-1}\frac{1}{\sqrt{\frac{k}{n}\left(1-\frac{k}{n}\right)}} =\frac{1}{2\pi} \int\limits_0^1\frac{dx}{\sqrt{x(1-x)}}=\frac{\Gamma(\frac{1}{2})^2}{2\pi~\Gamma(1)} = \frac{1}{2}$
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}&\color{#00f}{
\lim_{n \to \infty}\bracks{\expo{-n}\sum_{k = 0}^{n}{n^{k} \over k!}}}
\\[3mm]&=\lim_{n \to \infty}\bracks{\expo{-n}\sum_{k = 0}^{n}
\exp\pars{k\ln\pars{n} - \ln\pars{k!}}}
\\[3mm]&=
\lim_{n \to \infty}\braces{\expo{-n}\sum_{k = 0}^{n}
\exp\pars{n\ln\pars{n} - \ln\pars{n!} - {1 \over 2n}\bracks{k - n}^{2}}}
\\[3mm]&=
\lim_{n \to \infty}\braces{\expo{-n}\,{n^{n} \over n!}\sum_{k = 0}^{n}
\exp\pars{-{1 \over 2n}\bracks{k - n}^{2}}}
\\[3mm]&=
\lim_{n \to \infty}\braces{{\expo{-n}n^{n} \over n!}\int_{0}^{n}
\exp\pars{-{1 \over 2n}\bracks{k - n}^{2}}\,\dd k}
\\[3mm]&=
\lim_{n \to \infty}\bracks{{\expo{-n}n^{n} \over n!}\int_{-n}^{0}
\exp\pars{-\,{k^{2} \over 2n}}\,\dd k}
=
\lim_{n \to \infty}\bracks{{\expo{-n}n^{n} \over n!}\,\root{2n}
\int_{-\root{n}/2}^{0}\exp\pars{-k^{2}}\,\dd k}
\\[3mm]&=
\lim_{n \to \infty}\bracks{{\root{2}n^{n + 1/2}\expo{-n} \over n!}
\int_{-\infty}^{0}\exp\pars{-k^{2}}\,\dd k}
=
\lim_{n \to \infty}\bracks{{\root{2}n^{n + 1/2}\expo{-n} \over n!}
\,{\root{\pi} \over 2}}
\\[3mm]&=
\half\,\lim_{n \to \infty}\bracks{{\root{2\pi}n^{n + 1/2}\expo{-n} \over n!}}
=\color{#00f}{\Large\half}
\end{align}
A: If you'd like to see formal solution using calculus methods check this article http://www.emis.de/journals/AMAPN/vol15/voros.pdf
A: The sum is related to the partial exponential sum, and thus to the incomplete gamma function,
$$\begin{eqnarray*}
e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!}
&=& e^{-n} e_n(n) \\
&=& \frac{\Gamma(n+1,n)}{\Gamma(n+1)},
\end{eqnarray*}$$
since $e_n(x) = \sum_{k=0}^n x^k/k! = e^x \Gamma(n+1,x)/\Gamma(n+1)$.
But 
$$\begin{eqnarray*}
\Gamma(n+1,n) &=& \sqrt{2\pi}\, n^{n+1/2}e^{-n}\left(\frac{1}{2} + \frac{1}{3}\sqrt{\frac{2}{n\pi}} + O\left(\frac{1}{n}\right) \right).
\end{eqnarray*}$$
The first term in the asymptotic expansion for $\Gamma(n+1,n)$ can be found by applying the saddle point method to 
$$\Gamma(n+1,n) = \int_n^\infty dt\, t^n e^{-t}.$$
The higher order terms are in principle straightforward to compute. 
Using Stirling's approximation, we find 
$$e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!} 
= \frac{1}{2} 
+ \frac{1}{3}\sqrt{\frac{2}{n\pi}} 
+ O\left(\frac{1}{n}\right).$$
Thus, the limit is $1/2$, as found by @sos440 and @robjohn. 
This limit is a special case of DLMF 8.11.13. 
I just noticed a comment that suggests this be done using high school level math.
If this is a standard exercise at your high school, maybe they covered the incomplete gamma function! ;-)
A: 
The probabilistic way: 

This is $P[N_n\leqslant n]$ where $N_n$ is a random variable with Poisson distribution of parameter $n$. Hence each $N_n$ is distributed like $X_1+\cdots+X_n$ where the random variables $(X_k)$ are independent and identically distributed with Poisson distribution of parameter $1$. 
By the central limit theorem, $Y_n=\frac1{\sqrt{n}}(X_1+\cdots+X_n-n)$ converges in distribution to a standard normal random variable $Z$, in particular, $P[Y_n\leqslant 0]\to P[Z\leqslant0]$. 
Finally, $P[Z\leqslant0]=\frac12$ and $[N_n\leqslant n]=[Y_n\leqslant 0]$ hence $P[N_n\leqslant n]\to\frac12$, QED.

The analytical way, completing your try:

Hence, I know that what I need to do is to find $\lim\limits_{n\to\infty}I_n$, where
  $$
I_n=\frac{e^{-n}}{n!}\int_{0}^n (n-t)^ne^tdt.$$

To begin with, let $u(t)=(1-t)e^t$, then $I_n=\dfrac{e^{-n}n^n}{n!}nJ_n$ with
$$
J_n=\int_{0}^1 u(t)^n\mathrm dt.
$$
Now, $u(t)\leqslant\mathrm e^{-t^2/2}$ hence
$$
J_n\leqslant\int_0^1\mathrm e^{-nt^2/2}\mathrm dt\leqslant\int_0^\infty\mathrm e^{-nt^2/2}\mathrm dt=\sqrt{\frac{\pi}{2n}}.
$$
Likewise, the function $t\mapsto u(t)\mathrm e^{t^2/2}$ is decreasing on $t\geqslant0$ hence $u(t)\geqslant c_n\mathrm e^{-t^2/2}$ on $t\leqslant1/n^{1/4}$, with $c_n=u(1/n^{1/4})\mathrm e^{-1/(2\sqrt{n})}$, hence
$$
J_n\geqslant c_n\int_0^{1/n^{1/4}}\mathrm e^{-nt^2/2}\mathrm dt=\frac{c_n}{\sqrt{n}}\int_0^{n^{1/4}}\mathrm e^{-t^2/2}\mathrm dt=\frac{c_n}{\sqrt{n}}\sqrt{\frac{\pi}{2}}(1+o(1)).
$$
Since $c_n\to1$, all this proves that $\sqrt{n}J_n\to\sqrt{\frac\pi2}$. Stirling formula shows that the prefactor $\frac{e^{-n}n^n}{n!}$ is equivalent to $\frac1{\sqrt{2\pi n}}$. Regrouping everything, one sees that $I_n\sim\frac1{\sqrt{2\pi n}}n\sqrt{\frac\pi{2n}}=\frac12$.

Moral:
  The probabilistic way is shorter, easier, more illuminating, and more fun.
Caveat:
  My advice in these matters is, clearly, horribly biased.

A: Edited. I justified the application of the dominated convergence theorem.
By a simple calculation,
$$ \begin{align*}
e^{-n}\sum_{k=0}^{n} \frac{n^k}{k!}
&= \frac{e^{-n}}{n!} \sum_{k=0}^{n}\binom{n}{k} n^k (n-k)! \\
(1) \cdots \quad &= \frac{e^{-n}}{n!} \sum_{k=0}^{n}\binom{n}{k} n^k \int_{0}^{\infty} t^{n-k}e^{-t} \, dt\\
&= \frac{e^{-n}}{n!} \int_{0}^{\infty} (n+t)^{n}e^{-t} \, dt \\
(2) \cdots \quad &= \frac{1}{n!} \int_{n}^{\infty} t^{n}e^{-t} \, dt \\
&= 1 - \frac{1}{n!} \int_{0}^{n} t^{n}e^{-t} \, dt \\
(3) \cdots \quad &= 1 - \frac{\sqrt{n} (n/e)^n}{n!} \int_{0}^{\sqrt{n}} \left(1 - \frac{u}{\sqrt{n}} \right)^{n}e^{\sqrt{n}u} \, du.
\end{align*}$$
We remark that


*

*In $\text{(1)}$, we utilized the famous formula $ n! = \int_{0}^{\infty} t^n e^{-t} \, dt$.

*In $\text{(2)}$, the substitution $t + n \mapsto t$ is used.

*In $\text{(3)}$, the substitution $t = n - \sqrt{n}u$ is used.


Then in view of the Stirling's formula, it suffices to show that
$$\int_{0}^{\sqrt{n}} \left(1 - \frac{u}{\sqrt{n}} \right)^{n}e^{\sqrt{n}u} \, du \xrightarrow{n\to\infty} \sqrt{\frac{\pi}{2}}.$$
The idea is to introduce the function
$$ g_n (u) = \left(1 - \frac{u}{\sqrt{n}} \right)^{n}e^{\sqrt{n}u} \mathbf{1}_{(0, \sqrt{n})}(u) $$
and apply pointwise limit to the integrand as $n \to \infty$. This is justified once we find a dominating function for the sequence $(g_n)$. But notice that if $0 < u < \sqrt{n}$, then
$$ \log g_n (u)
= n \log \left(1 - \frac{u}{\sqrt{n}} \right) + \sqrt{n} u
= -\frac{u^2}{2} - \frac{u^3}{3\sqrt{n}} - \frac{u^4}{4n} - \cdots \leq -\frac{u^2}{2}. $$
From this we have $g_n (u) \leq e^{-u^2 /2}$ for all $n$ and $g_n (u) \to e^{-u^2 / 2}$ as $n \to \infty$. Therefore by dominated convergence theorem and Gaussian integral,
$$ \int_{0}^{\sqrt{n}} \left(1 - \frac{u}{\sqrt{n}} \right)^{n}e^{\sqrt{n}u} \, du = \int_{0}^{\infty} g_n (u) \, du \xrightarrow{n\to\infty} \int_{0}^{\infty} e^{-u^2/2} \, du = \sqrt{\frac{\pi}{2}}. $$
A: I do not know how much this will help you.
For a given $n$, the result is $\dfrac{\Gamma(n+1,n)}{n\ \Gamma(n)}$ which has a limit equal to $\dfrac12$ as $n\to\infty$.
