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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.

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10 Answers 10

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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

  1. In $\text{(1)}$, we utilized the famous formula $ n! = \int_{0}^{\infty} t^n e^{-t} \, dt$.
  2. In $\text{(2)}$, the substitution $t + n \mapsto t$ is used.
  3. 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,

$$ \int_{0}^{\sqrt{n}} \left(1 - \frac{u}{\sqrt{n}} \right)^{n}e^{\sqrt{n}u} \, du = \int_{0}^{\infty} g_n (u) \, du \xrightarrow{n\to\infty} \int_{0}^{\infty} e^{-u^2/2} \, du = \sqrt{\frac{\pi}{2}}. $$

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    $\begingroup$ Your second equation is closely related to this question (which I answered). $\endgroup$ – robjohn Jun 19 '12 at 15:47
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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.

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    $\begingroup$ One could simply say that for natural $n$ the median for the Poisson variable is $n$ (which, I believe, is nontrivial), and thus $P[N_n<n]=0.5$ immediately. $\endgroup$ – user35953 May 18 at 18:48
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    $\begingroup$ @Did, you are the reason we don't have Nobel prize in mathematics: they are too envious) $\endgroup$ – Alex Jul 21 at 10:25
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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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    $\begingroup$ Would the downvoter care to comment (he asked, expecting the answer "no")? $\endgroup$ – robjohn May 12 '18 at 14:01
  • $\begingroup$ hey rob, sadly the link in your answer is dead :( $\endgroup$ – tired Jun 10 '18 at 14:17
  • $\begingroup$ Yeah too many downvoters here :/ I upvoted this. Very Nice answer. $\endgroup$ – mick Jun 29 '18 at 21:49
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    $\begingroup$ The asymptotic approximation in $(11)$ can be improved by extending the approximations to get $$ \frac12+\frac1{\sqrt{2\pi n}}\left(\frac23-\frac{23}{270n}+\frac{23}{3024n^2}+\frac{259}{77760n^3}+O\!\left(\frac1{n^4}\right)\right) $$ and the big-$O$ term is close to $-\frac1{900n^4}$. $\endgroup$ – robjohn Mar 17 at 18:42
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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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    $\begingroup$ I wish you had copied the article here. It seemed small enough. References have a habit of pointing to NULL over time. $\endgroup$ – steven gregory Aug 22 '15 at 5:18
  • $\begingroup$ Please try to describe as much here as possible in order to make the answer self-contained. Links are fine as support, but they can go stale and then an answer which is nothing more than a link loses its value. Please read this post. $\endgroup$ – robjohn Jun 20 '19 at 3:14
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert #1 \right\vert} \newcommand{\yy}{\Longleftrightarrow}$ \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}

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    $\begingroup$ The second equal sign is a complete mystery. The passage from a sum to an integral also needs justification but it probably holds. $\endgroup$ – Did Mar 31 '14 at 16:25
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    $\begingroup$ Would you care answering @Did 's question regarding the second equal sign? I am miffed as well. Thank you, Felix Marin. $\endgroup$ – Hans May 11 '15 at 17:35
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    $\begingroup$ @Did's question has to be answered otherwise this seems to be an suitable answer $\endgroup$ – tired Oct 4 '15 at 12:12
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    $\begingroup$ yeah for me too, can you please answer @Did 's question? $\endgroup$ – user153330 Mar 1 '16 at 20:40
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    $\begingroup$ Is this supposed to justify the various unsubstantiated claims this post relies on? It does not. $\endgroup$ – Did Feb 8 '18 at 17:56
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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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    $\begingroup$ (+1) I derived this asymptotic expansion here in answer to a question on sci.math. I computed a few terms past $\frac12$: $$ \frac12+\frac{1}{\sqrt{2\pi n}}\left(\frac23-\frac{23}{270n}+\frac{23}{3024n^2}+\dots\right) $$ $\endgroup$ – robjohn Jun 20 '12 at 3:14
  • $\begingroup$ @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). $\endgroup$ – user26872 Jun 20 '12 at 3:47
  • $\begingroup$ @robjohn As steven gregory says above "I wish you had copied the article here. It seemed small enough. References have a habit of pointing to NULL over time" . Please, could you upload somewhere again? I'm curious about it. $\endgroup$ – vesszabo Jul 4 '17 at 21:13
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I'll give you two hints:

1) Poisson distributions;

2) Central limit theorem

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

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    $\begingroup$ 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? $\endgroup$ – DonAntonio Jun 19 '12 at 10:17
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    $\begingroup$ 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. $\endgroup$ – D. Thomine Jun 19 '12 at 10:43
  • $\begingroup$ The solution using Poisson distribution was also given here: Limit using Poisson distribution $\endgroup$ – Martin Sleziak Jun 19 '12 at 15:44
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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 $n\to\infty$.

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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)))}} $

$\displaystyle = \lim\limits_{n\to\infty} \frac{1}{2\pi}\frac{1}{n}\sum\limits_{k=1}^{n-1}\frac{1}{\sqrt{\frac{k}{n}\left(1-\frac{k}{n}\right)}} =\frac{1}{2\pi} \int\limits_0^1\frac{dx}{\sqrt{x(1-x)}}=\frac{\Gamma(\frac{1}{2})^2}{2\pi~\Gamma(1)} = \frac{1}{2}$

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  • $\begingroup$ (+1) Interesting idea. $\endgroup$ – user 1591719 Nov 26 '18 at 22:23
  • $\begingroup$ @user1357113 : That's kind of you, thanks. :) $\endgroup$ – user90369 Nov 27 '18 at 8:09
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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

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}e^{-n}\sum_{k=0}^{\lfloor Cn\rfloor}\frac{n^k}{k!}=\begin{cases}1&,C>1\\\\\frac12&, C=1\\\\0&, C<1\end{cases}}$$

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  • $\begingroup$ This is clever and I wonder if you could take it further: if $\lim\limits_{n\to\infty}e^{-n}\sum_{k=0}^{n+g_C(n)} = C$, can you characterize $g_C(n)$ at all, along the lines of things like the critical-value results for connectivity in random graph models? $\endgroup$ – Steven Stadnicki Sep 23 at 23:06
  • $\begingroup$ @StevenStadnicki First, thank you! Much appreciated. And yes, I thought about further generalizations, similar to the one you suggest, which should be fairly straightforward. $\endgroup$ – Mark Viola Sep 23 at 23:18

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