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Let $\{a_n\}_{n\in\mathbb{N}}$ be an increasing sequence of natural numbers, and $$ f_A(x)=\sum_{n\in\mathbb{N}}\frac{x^{a_n}}{a_n!}. $$ There are some cases in which the limit $$ l_A=\lim_{x\to+\infty} \frac{1}{x}\,\log(f_A(x)) $$ does not exist. However, if $\{a_n\}_{n\in\mathbb{N}}$ is an arithmetic progression, we have $l_A=1$ (it follows from a straightforward application of the discrete Fourier transform). Consider now the case $a_n=n^2.$

  • Is it true that there exists a positive constant $c$ for which $$\forall x>0,\quad e^{-x}f_A(x)=\sum_{k\in\mathbb{N}}x^k\left(\sum_{0\leq j\leq\sqrt{k}}\frac{(-1)^{k-j^2}}{(j^2)!\,(k-j^2)!}\right)\geq c\;?$$

  • Is it true that $l_A=1$?

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It is plain that $$\limsup_{x\to\infty} \frac{1}{x} \log f_{A}(x) = 1 $$ for any such sequence $A = (a_n)$, but I have no idea how to get a bound for liminf. – Sangchul Lee Nov 16 '12 at 12:04
Both numerical calculation and heuristic argument shows that $$\lim_{x\to\infty} e^{-x}f_A (x) = 0,$$ and also another numerical calculation suggests that $$ \lim_{x\to\infty} \frac{\log f_A (x)}{x} = 1. $$ – Sangchul Lee Nov 16 '12 at 13:18
up vote 2 down vote accepted

It is true that $l_A=1$. The logic is similar to my answer to $\lim\limits_{x\to\infty}f(x)^{1/x}$ where $f(x)=\sum\limits_{k=0}^{\infty}\cfrac{x^{a_k}}{a_k!}$. Firstly, the terms after $n=3x$ don't matter:

$$\sum_{n=3x}^\infty x^n/n! <\sum_{n=3x}^\infty x^n /(3x/e)^n=C$$ by Stirling approximation.

But for $n<3x$ there is going to be a perfect square $s$ between $n$ and $n+2\sqrt{3x}$ (this is just $(k+1)^2-k^2=2k+1$ and $k< \sqrt{3x}$). Then the values $x^s/s!$ and $x^n/n!$ differ by a factor of at most $(3x)^{2\sqrt{3x}}$,

So if I multiply each term $x^s/s!$ by that ratio and take at least $2\sqrt{3x}$ copies I will have for each $x^n/n!$ (with $n<3x$) at least one term at least as big. This means that $$f_A(x) (3x)^{2\sqrt{3x}}{2\sqrt{3x}} +C > e^x$$ and so $l_A=1$.

(I think the ratio of terms can actually be made $3^{2\sqrt{3x}}$, but it works as is too.)

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I have not checked all the detail yet (I could not sleep at all this night!), but the tail-part estimation in the first line is already fascinating. I love this answer, and of course +1 upvote! – Sangchul Lee Nov 18 '12 at 21:37
$$\left(\sum_{m^2<3x}\frac{x^{m^2}}{(m^2)!}\right)\cdot\left(1+x+\ldots+x^{2 \lfloor\sqrt{3x}\rfloor}\right)\geq\sum_{n\leq 3x}\frac{x^n}{n!}\geq e^x-C,$$ simple and clever. I also appreciate the fact that this argument admits a straightforward generalization for all the power-like sequences $a_n=n^k$, award incoming. – Jack D'Aurizio Nov 19 '12 at 10:32
@JackD'Aurizio Yes, that's a better way to write it. – Max Nov 19 '12 at 16:25

This is an answer to the first question only, for now.

There does not exist such a constant $c>0$. Assume it does, i.e., that $e^{-x} f_{A}(x) \ge c > 0$ for all $x>0$. Define $f_{A,0} = f_A$, and inductively $$ f_{A,{k}}(x) = c+ \int_0^x f_{A,k-1}(t) \, dt = c \sum_{j=0}^{k-1} \frac{x^j}{j!} + \sum_{n \in \mathbb{N}} \frac{x^{a_n+k}}{(a_n+k)!} $$ for $k \ge 1$. (The power series representation is easily checked by induction.) By assumption we have $f_{A,0}(x) \ge c e^x$, and then the induction step $$ f_{A,k}(x) \ge c + \int_0^x ce^t \, dt = c+ ce^x-c = ce^x $$ shows that $f_{A,k}(x) \ge c e^x$ for all $k \ge 0$ and $x>0$.

Now if $(a_n)$ is any sequence with $\lim\limits_{n\to\infty} (a_{n+1}-a_n) = \infty$, such as your example $a_n = n^2$, the power series representation for $f_{A,k}$ shows that $$ F_{A,m}(x) = \sum_{k=0}^{m-1} f_{A,k}(x) \le P_{A,m}(x) + e^x $$ where $P_{A,m}$ is a polynomial. Then $$ P_{A,m}(x) + e^x \ge \sum_{k=0}^{m-1} ce^x = mc e^x $$ for all $x>0$. This implies $mc \le 1$, so $c \le \frac1m$. However, as $m$ was arbitrary, this contradicts $c>0$.

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There is an error in the argument. See the comments.

It is a nice exercise to show $l_A = 1$ for every polynomial $n^k$. Let me sketch the proof for $k = 2$. Observe that for $x \geq 0$: $$\sum_{n=0} \frac{x^{n^2}}{(n^2)!} \leq \sum_{n=0} \frac{x^n}{n!} = e^x$$ simply, because the left hand side is a subsequence of the right hand side. The next thing to do is to bound our sequence from the left. We have: $$\sum_{n=0} (2n + 1)\frac{x^{(n+1)^2}}{(n+2)^2!} \geq \sum_{n=0}\frac{x^n}{(n+2)!}$$ because in each group of the size $(n+1)^2 - n^2 = 2n + 1$ the last nominator is the biggest and the first denominator is the smallest one. For sufficiently large $x$ (for example $x \geq 2$): $$\sum_{n=0} \frac{x^{(n+2)^2}}{(n+2)^2!} \geq \sum_{n=0} (2n+1)\frac{x^{(n+1)^2}}{(n+2)^2!}$$ thus $$\sum_{n=0} \frac{x^{n^2}}{(n^2)!} - x - 1= \sum_{n=2} \frac{x^{n^2}}{(n^2)!} \geq \sum_{n=2} \frac{x^{(n-2)}}{n!} = \frac{1}{x^2}\sum_{n=2} \frac{x^n}{n!} = \frac{1}{x^2}(e^x - x - 1)$$ Therefore, for sufficiently large $x$: $$\frac{e^x}{x^2} \leq \frac{e^x}{x^2} + x + 1 - \frac{x+1}{x^2} \leq \sum \frac{x^{n^2}}{(n^2)!} \leq e^x$$

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I have some serious trouble in the understanding of your proof. May you write the summation ranges all the time? Anyway, I am quite skeptical, since your approach seems to produce effective lower bounds for any increasing sequence $\{a_n\}_{n\in\mathbb{N}}$, and I know for sure that the limit $\lim_{x\to+\infty}\frac{\log f_A(x)}{x}$ does not always exist. – Jack D'Aurizio Nov 16 '12 at 16:32
It does not. Try to imitate the argument for $a_n = n^n$. – Michal R. Przybylek Nov 16 '12 at 17:08
About the second formula in your proof, what I get is: $$\sum_{n=0}^{+\infty}\frac{x^n}{(n+2)!}=\sum_{k=0}^{+\infty}\sum_{n=k^2}^{(k+1)‌​^2-1}\frac{x^n}{(n+2)!}\leq\sum_{k=0}^{+\infty}(2k+1)\frac{x^{(k+1)^2}}{(k^2+2)!}‌​,$$ that is slighly different from what you wrote. Did you use a different "regrouping" technique? – Jack D'Aurizio Nov 16 '12 at 17:40
I am sorry, I have made an error here. Right now, I am not even sure if the whole argument is fixeable. Thnks for spotting the bug. – Michal R. Przybylek Nov 16 '12 at 18:28

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