# Existence of $A$ such that $\lim_{x\to\infty}\operatorname{poly}(x) e^{-x} \sum_{n\in A} \frac{x^n}{n!}=1$

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.

• 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. Jun 21, 2020 at 18:02
• 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... Jun 21, 2020 at 18:15
• What methods have you tried to find the structure of $A$? Could you show your progress explicitly? Jun 23, 2020 at 14:49

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 $$x \in A$$.

• +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). Jun 24, 2020 at 10:13
• @EwanDelanoy um... i literally just gave an infinite sequence along which $f$ goes to infinity. there's really nothing going on 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:

• Hi. reading through your answer now. u agree there are more cases than case $1$ and case $2$, right? Jun 28, 2020 at 1:29
• @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. Jun 28, 2020 at 1:34
• where does $1/6$ come in to play? Jun 28, 2020 at 6:48
• @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 $\alpha \in (0, \frac{1}{2})$. Jun 28, 2020 at 7:05
• 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! 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: $$$$h(x):=e^{-x} \sum_{n\in A} \dfrac{x^n}{n!}$$$$ Then for every $$m\in A$$ we have $$$$h(m)=e^{-m} \sum_{n\in A} \dfrac{m^n}{n!} \geq e^{-m} \dfrac{m^m}{m!}$$$$ We know $$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$$ we get: $$$$\limsup_{x\to + \infty} |p(x)h(x)| = +\infty$$$$ This shows that a polynomial $$q(x)$$ satisfying: $$$$\lim_{x\to\infty} q(x) e^{-x} \sum_{n\in A} \dfrac{x^n}{n!}=1$$$$ 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: $$$$\lim_{x\to +\infty} e^{-x} \sum_{n\in A} \dfrac{x^n}{n!}=C$$$$ ?

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:

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

• 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? Jun 27, 2020 at 22:12
• @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? Jun 27, 2020 at 22:48
• 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. Jun 27, 2020 at 22:51