# Evaluating $\lim_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$

I'm supposed to calculate:

$$\lim_{n\to\infty} e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!}$$

By using W|A, i may guess that the limit is $\frac{1}{2}$ that is a pretty interesting and nice result. I wonder in which ways we may approach it.

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How do you formulate this for W|A to get that result? I didn't get it work... –  Gottfried Helms Jun 19 '12 at 9:46
@Gottfried Helms: wolframalpha.com/input/?i=%28sum_k%3D0^500+%28%28500^k%29%2Fk!%29*%28‌​e^%28-500%29%29. It seems the limit goes to $\frac{1}{2}$. –  Chris's sis Jun 19 '12 at 9:47
hmm, for me W|A says zero for your expression... (where we have the constant 500, why? you can ask for limits, too) What's going on here? –  Gottfried Helms Jun 19 '12 at 9:51
@GottfriedHelms Try this and this. :-) –  dtldarek Jun 19 '12 at 10:02
–  leonbloy Jun 19 '12 at 15:46

If you'd like to see formal solution using calculus methods check this article http://www.emis.de/journals/AMAPN/vol15/voros.pdf

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By simple(?) manipulation,

\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)! \\ &= \frac{e^{-n}}{n!} \sum_{k=0}^{n}\binom{n}{k}\int_{0}^{\infty} n^k t^{n-k}e^{-t}\;dt\\ &= \frac{e^{-n}}{n!} \int_{0}^{\infty} (n+t)^{n}e^{-t}\;dt\\ &= \frac{n^n e^{-n}}{n!} \int_{0}^{\infty} \left(1+\frac{t}{n}\right)^{n}e^{-t}\;dt. \end{align*}

Thus from Stirling's formula, it suffices to show that

$$\frac{1}{\sqrt{2\pi n}} \int_{0}^{\infty} \left(1+\frac{t}{n}\right)^{n}e^{-t}\;dt \xrightarrow{n\to\infty} \frac{1}{2}.$$

Remark that

1. $2^x \geq 1+x$ for all $x\geq 1$, and
2. we have $$x - \frac{x^2}{2} \leq \log (1+x) = x - \frac{x^2}{2} +O(x^3) \leq x - \frac{x^2}{6}$$ for $0\leq x \leq 1$.

From this, we first note that

$$\frac{1}{\sqrt{2\pi n}} \int_{n}^{\infty} \left(1+\frac{t}{n}\right)^{n}e^{-t}\;dt \leq \frac{1}{\sqrt{2\pi n}} \int_{n}^{\infty} (e/2)^{-t}\;dt \xrightarrow{n\to\infty} 0.$$

Moreover,

\begin{align*} \frac{1}{\sqrt{2\pi n}} \int_{0}^{n} \left(1+\frac{t}{n}\right)^{n}e^{-t}\;dt &= \frac{1}{\sqrt{2\pi n}} \int_{0}^{n} \exp \left(-\frac{t^2}{2n} + O\left(\frac{t^3}{n^2}\right)\right)\;dt \\ &= \frac{1}{\sqrt{2\pi}} \int_{0}^{\sqrt{n}} \exp \left(-\frac{u^2}{2} + O\left(\frac{u^3}{\sqrt{n}}\right)\right)\;du \quad (t = u\sqrt{n}), \end{align*}

Which converges to

$$\frac{1}{\sqrt{2\pi}} \int_{0}^{\infty} e^{-u^2/2}\;du = \frac{1}{2}$$

by dominated convergence theorem and Gaussian integral. Therefore the limit follows by our observations.

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Wow! +1 ${}{}{}{}{}{}$ –  Sam Jun 19 '12 at 14:14
Your second equation is closely related to this question (which I answered). –  robjohn Jun 19 '12 at 15:47

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 (as one can see) horribly biased.

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Thanks. Deleting. –  Pedro Tamaroff Sep 14 '13 at 15:44
+1 your moral and the caveat. –  Lost1 Dec 30 '13 at 23:55

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}

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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! ;-)

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(+1) I derived this asymptotic expansion here in answer to a question on sci.math. I computed a few terms past $\frac12$: $$\frac12+\frac{1}{\sqrt{2\pi n}}\left(\frac23-\frac{23}{270n}+\frac{23}{3024n^2}+\dots\right)$$ –  robjohn Jun 20 '12 at 3:14
@robjohn: Thanks for the link, I'll have a look. By the way, I voted up your nice solution a couple of hours ago. I like your short and sweet derivation of (3). –  user26872 Jun 20 '12 at 3:47
@Chris: Glad to help. And if you find an elementary proof first, make sure to post it! :-) –  user26872 Jun 20 '12 at 5:59

I'll give you two hints:

I am not aware of any other technique to solve the problem, so any other answer is appreciated.

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In what country's education system is this a high school exercise?! I ask sincerely because I'm interested in that, and my experience is that even some of the most advanced mathematically education systems don't reach exercises as the one you're proposing...not even close. Of course, I don't know all the education systems (not even close, again), but the expression within the sum seems to be pretty tough...it reminds the series for the exponential function, but that n repeating in the numerator and upper limit...are you sure the expression is accurate? –  DonAntonio Jun 19 '12 at 10:17
This problem is definitely not at high-school level. Without the hints I gave, I doubt most university students (even graduates) would solve it in reasonable time. –  D. Thomine Jun 19 '12 at 10:43
@Chris, thank you for checking. As D. Thomine says, this seems to be ("seems" because I accept the possibility someone else come up with some trick that'll make the solution to your problem embarrasingly easy...) a rather tough exercise, so if you're a high school student I think you shouldn't be too hard on yourself if you are having a rough time tackling this little beast, even with the given hints, which definitely are university level. –  DonAntonio Jun 19 '12 at 10:48
The solution using Poisson distribution was also given here: Limit using Poisson distribution –  Martin Sleziak Jun 19 '12 at 15:44

For a given $n$, the result is $\dfrac{\Gamma(n+1,n)}{n\ \Gamma(n)}$ which has a limit equal to $\dfrac12$ as $n\to\infty$.
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ \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}