I want to know if there exists a set $A \subseteq \mathbb{N}$ such that $$ \lim_{x\to\infty} x^2 e^{-x} \sum_{n\in A} \dfrac{x^n}{n!}=1 $$

More generally, the question will be the existence of a set $A$ that $$ \lim_{x\to\infty}\operatorname{poly}(x) e^{-x} \sum_{n\in A} \dfrac{x^n}{n!}=1 $$

When $A$ is finite, it is obvious that the limit must be $0$. But when $A$ is infinite, the structure of $A$ can be very complex, and I don't know how to proceed further.

  • $\begingroup$ Great question! I don't think $A$ can contain arithmetic progressions, else the series looks like a multisection of $e^x$, one term of which is a factor of $e^x$ which is not cancelled in the limit. $\endgroup$
    – Integrand
    Jun 21, 2020 at 18:02
  • $\begingroup$ Note that if $g(x)=(1-x)^{-1}$ is the sum of the geometric series, for $|x|<1$, if $\text{poly}(x) = a_d x^d +\cdots +a_0$, we have $\lim_{x\to\infty} \text{poly}(x) \cdot 1/ g(x) \cdot g(x^{d+1}) = a_d$. This is because the geometric series sums to a rational function, which behaves better than exponentials here... $\endgroup$
    – Integrand
    Jun 21, 2020 at 18:15
  • $\begingroup$ What methods have you tried to find the structure of $A$? Could you show your progress explicitly? $\endgroup$ Jun 23, 2020 at 14:49

3 Answers 3


No such $A$ exists. Clearly such an $A$ would have to be infinite. Note that $m^2e^{-m}\frac{m^m}{m!} \sim m^2e^{-m}\frac{m^m}{\frac{m^m}{e^m}\sqrt{2\pi m}} = \frac{1}{\sqrt{2\pi}}m^{3/2}$, so restricting to $x=m \in A$ and just looking at the term $n=m$ shows the limit is infinity along $x \in A$.

  • 3
    $\begingroup$ +1, nice proof, but I don't like your presentation, I had to rewrite it entirely on my own to convince myself it's correct. Here's how I would have put it : let $f(x)=x^2e^{-x}\sum_{n\in A}\frac{x^n}{n!}$. As you said $A$ is infinite, put $A=\lbrace a_k\rbrace_{k\geq 1}$ where $(a_k)$ is increasing. As you explained, $f(a_k)\geq g(a_k)$ where $g(m)=m^2e^{-m}\frac{m^m}{m!}$. Since $\lim_{+\infty}g=+\infty$, we cannot have $\lim_{+\infty}f=1$ (note that we haven't shown that $\lim_{+\infty}f=+\infty$ though). $\endgroup$ Jun 24, 2020 at 10:13
  • $\begingroup$ @EwanDelanoy um... i literally just gave an infinite sequence along which $f$ goes to infinity. there's really nothing going on $\endgroup$ Jun 24, 2020 at 14:30

For each subset $A$ of $\mathbb{N}_0 = \{0, 1, 2, \dots\}$, we define

$$ f_A(x) := \sum_{n \in A} \frac{x^n}{n!}. $$

1. @mathworker21's proof essentially shows that, for any infinite subset $A$ of $\mathbb{N}_0$,

$$ \limsup_{x\to\infty} \sqrt{x}e^{-x}f_A(x) \geq \frac{1}{\sqrt{2\pi}}. $$

So, for any non-constant polynomial $p(x)$, we must have

$$ \limsup_{x\to\infty} |p(x)|e^{-x}f_A(x) = \infty $$

and the OP's condition cannot be satisfied.

2. Based on the above observation, we may formulate another interesting question:

Question. Let $0 \leq \alpha \leq \frac{1}{2}$ and $\ell > 0$. Is there $A \subseteq \mathbb{N}_0$ such that $$ \lim_{x\to\infty} x^{\alpha} e^{-x}f_A(x) = \ell $$

Case 1. When $\alpha = 0$, we necessarily have $\ell \in (0, 1]$ for an obvious reason. Now we claim that any values of $\ell \in (0, 1]$ can be realized.

  • Let $m \geq 1$ and $R \subseteq \{0, 1, \dots, m-1\}$. Then $$ \lim_{x\to\infty} e^{-x} \sum_{q=0}^{\infty}\sum_{r\in R} \frac{x^{qm+r}}{(qm+r)!} = \frac{|R|}{m}. $$ This lemma is an easy consequence of the following explicit computation \begin{align*} \sum_{q=0}^{\infty} \frac{x^{qm+r}}{(qm+r)!} &= \frac{1}{m}\sum_{k=0}^{m-1} e^{-\frac{2\pi i k r}{m}} \exp\left(e^{\frac{2\pi i k}{m}}x\right) \\ &= \frac{e^x}{m} + \mathcal{O}\left(\exp\left(x \cos(\tfrac{2\pi}{m})\right)\right) \qquad\text{as}\quad x \to \infty. \end{align*} So, the case of rational $\ell$ is resolved.

  • When $\ell$ is irrational, define $A$ by as follows: Set $$ A_1 = \begin{cases} \{0\}, & \text{if $\ell \in (0,\frac{1}{2}]$}; \\ \{0,1\}, & \text{if $\ell \in (\frac{1}{2}, 1]$}. \end{cases} $$ Next, suppose that $A_k$ is defined and contains $\lceil 2^k \ell \rceil$ elements. Consider the set $A_k \cup (2^k + A_k)$. This set contains $2\lceil 2^k \ell \rceil$ elements. Then by removing its last element if necessary, reduce the number of its elements to $\lceil 2^{k+1}\ell \rceil$. Denote the resulting set by $A_{k+1}$. Finally, set $A = \cup_{k=1}^{\infty} (2^k + A_k)$. It can be shown that this set achieves the desired property.

Case 2. When $\alpha = \frac{1}{2}$, I suspect that no such $\ell$ exists. I have a couple of heuristic arguments for this guess, mainly based on the case $A = \{n^2 : n \in \mathbb{N}_0\}$. A heuristic computation suggests that

$$ \sqrt{x}e^{-x} \sum_{n=0}^{\infty} \frac{x^{n^2}}{(n^2)!} \sim \frac{1}{\sqrt{2\pi}} \sum_{k=-\infty}^{\infty} e^{-\frac{(2k-r)^2}{2}}, \qquad r = \frac{x-\lfloor\sqrt{x}\rfloor^2}{\sqrt{x}} $$


$$ \textbf{Figure.} \ \text{ A comparison of the left-hand side (blue) and the right-hand side (orange).}$$

which oscillates as $x\to\infty$. The main mechanism of this oscillatory behavior is that, if $x$ is large, then each term $\frac{x^n}{n!}$ with $n = x + \mathcal{O}(x^{1/2})$ will contribute to $\sqrt{x}e^{-x}f_A(x) $. I am currently trying to formalize this idea to prove my conjecture.

Case 3. When $0 < \alpha < \frac{1}{2}$, I propose the following conjecture:

  • Conjecture. Let $\beta = \frac{1}{1-\alpha}$ and $c > 0$, and define $A$ by $$ A = \{ \lfloor (cn)^{\beta} \rfloor : n \geq 0 \}. $$ Then $$ \lim_{x \to \infty} x^{\alpha} e^{-x} f_A(x) = \frac{1}{\beta c}. $$

For instance, the following example illustrates the case of $\alpha = \frac{1}{7}$ and $c = 3$:


$$ \textbf{Figure.} \ \text{ $x^{\alpha}e^{-x}f_A(x)$ (blue) and its limit $\frac{1}{\beta c}$ (orange) when $\alpha = \frac{1}{7}$ and $c = 3$}$$

For the remaining portion of this part, we sketch the proof of this conjecture when $0 < \alpha < \frac{1}{6}$. The main idea is that the sum can be truncated:

  • Lemma. Fix a function $\lambda = \lambda(x) \geq 0$ such that $\lambda \to \infty$ and $\frac{\lambda}{\sqrt{x}} \to 0$ as $x \to \infty$. Then there exists a constant $C > 0$, depending only on $\lambda$, such that $$ e^{-x} \sum_{|n - x| > \lambda\sqrt{x}} \frac{x^n}{n!} \leq \frac{C}{\lambda}. $$

Now we further assume that $\frac{\lambda}{x^{1/6}} \to 0$ as $x \to \infty$. Then using the above lemma, we can show:

$$ e^{-x}f_A(x) = \frac{1 + \mathcal{O}(\lambda^3/\sqrt{x})}{\sqrt{2\pi x}} \sum_{\substack{|m - x| \leq \lambda\sqrt{x} \\ m \in A}} e^{-\frac{(m-x)^2}{2x}} + \mathcal{O}\left(\frac{1}{\lambda}\right). $$

For each $m \in A$, let $n_m$ be defined by $m = \lfloor (c n_m)^{\beta} \rfloor$, and write $t_m = n_m - c^{-1}x^{1/\beta}$. Then we can show that, uniformly in $x$ and $m \in A \cap [x-\lambda\sqrt{x}, x+\lambda\sqrt{x}]$,

$$ \frac{(m-x)^2}{2x} = \frac{1}{2} \biggl( \frac{\beta c t_m}{x^{\frac{1}{2}-\alpha}} \biggr)^2 + o(1). $$

So, if in addition $\lambda$ is chosen that $x^{\alpha}/\lambda \to 0$ (which is possible by letting $\lambda(x) = x^{\gamma}$ for some $\gamma \in (\alpha, \frac{1}{6})$), then

$$ x^{\alpha} e^{-x}f_A(x) = \frac{1 + o(1)}{\sqrt{2\pi}} \sum_{\substack{|m - x| \leq \lambda\sqrt{x} \\ m \in A}} \exp\biggl[ - \frac{1}{2} \biggl( \frac{\beta c t_m}{x^{\frac{1}{2}-\alpha}} \biggr)^2 \biggr] \frac{1}{x^{\frac{1}{2}-\alpha}} + o(1), $$

which converges to

$$ \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-\frac{(\beta c u)^2}{2}} \, \mathrm{d}u = \frac{1}{\beta c} $$

as $x \to \infty$.

3. For different lines of questions regarding the asymptotic behavior of $f_A(x)$, see:

  • $\begingroup$ Hi. reading through your answer now. u agree there are more cases than case $1$ and case $2$, right? $\endgroup$ Jun 28, 2020 at 1:29
  • $\begingroup$ @mathworker21, Indeed. I am currently playing with the intermediate case $0 < \alpha < \frac{1}{2}$ to see what I can expect. I guess that we may find such $A$'s in this case, although I have not enough supporting evidences to this. $\endgroup$ Jun 28, 2020 at 1:34
  • $\begingroup$ where does $1/6$ come in to play? $\endgroup$ Jun 28, 2020 at 6:48
  • $\begingroup$ @mathworker21, It is rather a technical reason. The estimations used in my proof requires that both $\lambda/x^{1/6}$ and $x^{\alpha}/\lambda$ converges to $0$ as $x\to\infty$, which forces $\alpha < \frac{1}{6}$. But both the numerical experiments and my gut are telling that the conjecture should be true for all $\alpha \in (0, \frac{1}{2})$. $\endgroup$ Jun 28, 2020 at 7:05
  • $\begingroup$ By treating the case $\alpha=0$ you completely solved the OP question, maybe it's worth opening a new post for $\alpha>0$. I think it's amazing that for $\ell \in (0,1)$ you found $A$ satisfying precisely $\lim_n \lvert A \cap \{0,1,2,\dots n-1\}\rvert/n = \ell$. Wow! $\endgroup$ Jun 28, 2020 at 9:37

This is a partial answer to the more general case. I'll use mathworker21 argument again. Suppose $A$ is infinite. Define: \begin{equation} h(x):=e^{-x} \sum_{n\in A} \dfrac{x^n}{n!} \end{equation} Then for every $m\in A$ we have \begin{equation} h(m)=e^{-m} \sum_{n\in A} \dfrac{m^n}{n!} \geq e^{-m} \dfrac{m^m}{m!} \end{equation} We know $e^{-m} \dfrac{m^m}{m!}\sim \dfrac{1}{\sqrt{2\pi m}}$ as $m\to\infty$ thanks to Stirling formula. Let p(x) be a polynomial of degree $\geq 1$. Evaluating $p(x)h(x)$ along $A$ we get: \begin{equation} \limsup_{x\to + \infty} |p(x)h(x)| = +\infty \end{equation} This shows that a polynomial $q(x)$ satisfying: \begin{equation} \lim_{x\to\infty} q(x) e^{-x} \sum_{n\in A} \dfrac{x^n}{n!}=1 \end{equation} must be a constant polynomial.

EDIT. Following mathworker21 suggestion, I'll extend this answer. The general problem can be restated in what follows:

Given a constant $C>0$, does a set $A \subseteq \mathbb{N}$ exist satisfying: \begin{equation} \lim_{x\to +\infty} e^{-x} \sum_{n\in A} \dfrac{x^n}{n!}=C \end{equation} ?

I currently don't have a general answer. What one can say is that, since $\sum_{n\in A} \dfrac{x^n}{n!} \leq e^x$ such a $C$ must be $\leq 1$. More over, for some specific values of $C$ it is possible to build a corresponding $A$ satisfy the problem. For $N \in \mathbb{N}-\{0\}$ , set $\alpha:=e^{2\pi i/N}$, $A:=N\mathbb{N}$. We have:

\begin{equation} \sum_{n\in A} \dfrac{x^n}{n!} = \sum_{k=0}^{\infty} \dfrac{x^{kN}}{(kN)!} = \dfrac{\sum_{k=1}^{N} e^{\alpha^k x}}{N} \sim \dfrac{e^{x}}{N} \end{equation} as $x\to +\infty$. This shows the answer is yes for $C=1/N$.

  • $\begingroup$ Not to be rude, but I think all of this was already implicit in my answer. Also, doesn't taking $A$ to be the evens and $q(x) = 2$ work? $\endgroup$ Jun 27, 2020 at 22:12
  • $\begingroup$ @mathworker21 If the general problem is: "given a polynomial $q(x)$, is there a suitable $A$ satisfying $ \lim_{x\to\infty} q(x) e^{-x} \sum_{n\in A} \dfrac{x^n}{n!}=1$ ?" then I simply pointed out, exploiting your answer, that a necessary condition on $q$ in order to find a suitable $A$ is that $q$ must be constant. It's your idea though, maybe I should have commented your answer instead. Should I delete the answer? $\endgroup$ Jun 27, 2020 at 22:48
  • $\begingroup$ Yes I know. What I was saying is that it is pretty immediate that my answer implies $q$ must be constant. You can keep your answer if you want. Maybe you can try to add a proof that if $q(x)$ is a constant greater than $1$, then an $A$ does exist. For example, if $q(x) \equiv n$ an integer $n \ge 1$, then you can take $A$ to be all multiples of $n$. I think what I just said is true; I'm not $100\%$ sure. $\endgroup$ Jun 27, 2020 at 22:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.