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$\renewcommand{\ge}{\geqslant}\renewcommand{\le}{\leqslant}$ $\newcommand{\pa}[1]{\left( #1 \right)}$

Let us take $\alpha > 0$, $x_1 := \alpha$ and for any $n \ge \mathbb{N}$, \begin{align*} \boxed{x_n := \sum_{s=1}^{n-1} x_{n-s} x_s,} \end{align*} There exist $\xi,c \in \mathbb{R}_+$, such that for any $n \in \mathbb{N}$, \begin{equation} x_n \le c (\xi \alpha)^n n^{-\frac 32}. \qquad\qquad (EQ1) \end{equation}

Two series are equivalent if their quotient goes to $1$ as $n \rightarrow +\infty$. Numerical evidence suggests that \begin{align*} \boxed{x_n \underset{\substack{n \rightarrow +\infty}}{\sim} \; c \; \frac{(\xi \alpha)^n}{n^{\frac 32}}} \end{align*} with $\xi = 4$ and $c = \frac{1}{\xi \sqrt{\pi}}$. Would anyone know how to prove this numerical evidence please ? It's a "Cauchy square" in some sense so I guess the series has already been studied ? The proof of (EQ1) below gives an idea of the mechanisms. With the $S_n$ in the proof, numerical evidence suggests that $S_n \underset{\substack{n \rightarrow +\infty}}{\sim} 5.2247 / n^{\frac 32}$.


Remark : Take $\beta$. If we can prove the numerical conjecture, then we can deduce that if $y_1 = \alpha$ and $y_n = \beta \sum_{s=1}^{n-1} y_{n-s} y_s$, then $y_n \underset{\substack{n \rightarrow +\infty}}{\sim} \; c (\xi \alpha \beta)^n n^{-\frac 32}$ with $c = \frac{1}{\beta \xi \sqrt{\pi}}$, just by renormalization $y_n = x_n/\beta$. In other words the case $\beta = 1$ implies the general case. We could have normalized to make $\alpha = 1$, but we cannot normalize such that both $\alpha$ and $\beta$ are equal to 1.


Proof of (EQ1)

Let us prove (EQ1) inductively. Take $n \in \mathbb{N}$, $n \ge 2$, such that for any $s \in \{1,\dots,n-1\}$, $x_s \le c (\xi\alpha )^s s^{-\frac 32}$. Then \begin{align}\label{eq:bound_xn_interm} x_n \le c^2 \pa{\xi\alpha}^n \sum_{s=1}^{n-1} \frac{1}{\pa{s(n-s)}^{\frac 32}} = c^2 \pa{\xi\alpha}^n S_n, \end{align} where for any $x \in ]0,n[$ we define $g(x) := \pa{x(n-x)}^{-\frac 32}$ and want to estimate $S_n := \sum_{s=1}^{n-1} g(s)$ for any $n \ge 2$. By Carlo and Henri, we have $n^{3/2} S_n \rightarrow 2 \zeta(3/2) \simeq 5.2247$, we can also show easily that $S_n \le \frac{\xi}{n^{\frac 32}}$ where $\xi = 10$. Then we have \begin{align*} x_n \le ( \xi c) c (\xi \alpha)^n n^{-\frac 32}. \end{align*} We take $c = 1/\xi$ and the recursive equation is proved. Now for $n=1$ the right hand side of (EQ1) is $\alpha$ so the initial step is also valid.


Where does the "4" (of the numerical insight) could come from ? One possible starting point is presented here.

The function $g$ is decreasing on $]0,n/2]$, increasing on $[n/2,n[$, and for any $x \in ]0,n/2]$, $g(n-x) = g(x)$. By a series/integral comparision, we have \begin{align*} \int_1^{n-1} g \le S_n - g\pa{\frac n2}, \qquad S_n + g\pa{\frac n2} \le g(1) + g(n-1) + \int_1^{n-1} g \end{align*} when $n \in 2 \mathbb{N}$ and \begin{align*} \int_1^{n-1} g \le S_n - g\pa{\frac{n-1}{2}}, \qquad S_n + g\pa{\frac n2} \le g(1) + g(n-1) + \int_1^{n-1} g \end{align*} when $n \in 2 \mathbb{N} +1$, hence for all $n\in \mathbb{N}$ such that $n \ge 2$, \begin{align*} \int_1^{n-1} g \le S_n \le \frac{2}{(n-1)^{\frac 32}} + \int_1^{n-1} g. \end{align*} Defining $G(x) := \frac{n-2x}{\sqrt{x(n-x)}}$ we have $G'(x) = -\frac{n^2}{2} g(x)$ so \begin{align*} \int_1^{n-1} g = \frac{2}{n^2} \pa{G(1) - G(n-1)} = \frac{4 (n-2)}{n^2\sqrt{n-1}} \underset{\substack{n \rightarrow +\infty}}{\sim} \; \frac{4}{n^{\frac 32}}. \end{align*} For any $n \ge 2$, $\frac{n-2}{\sqrt{n-1}} \le \sqrt{n}$ so $\int_1^{n-1}g \le 4 n^{-\frac 32}$ and \begin{align*} \frac{4}{n^{\frac 32}} \pa{1 - \frac{2}{n} } \le S_n \le \frac{6}{n^{\frac 32}} \pa{1 + \frac{2}{n} }. \end{align*}

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  • $\begingroup$ $\gamma=1/\sqrt{\pi}$. $\endgroup$ Commented Apr 8 at 15:41
  • $\begingroup$ Indeed, verified to relative precision $10^{-3} $, thanks ! $\endgroup$
    – Raphaël
    Commented Apr 8 at 15:52

2 Answers 2

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You can reduce the evaluation of $S_n$ to a quadrature by means of the Abel-Plana formula, $$S_n=\sum _{s=1}^{n-1} g(s)=\int_1^{n-1}g(s)\,ds+\tfrac{1}{2}g(1)+\tfrac{1}{2}g(n-1)$$ $$\qquad\qquad -\,2\operatorname{Im}\int_0^\infty \frac{g(1+i y)-g(n-1+i y)}{e^{2 \pi y}-1}\,dy$$ $$\qquad=\frac{5n^2-12n+8}{(n-1)^{3/2} n^2}-4\operatorname{Im}\int_0^\infty \frac{g(1+i y)}{e^{2 \pi y}-1}\,dy,$$ where $g(s)=[s(n-s)]^{-3/2}=g(n-s)$.

The integral can be expanded in a power series in $n$, to leading order I find $$\lim_{n\rightarrow\infty} n^{3/2}S_n=5-4\operatorname{Im}\int_0^\infty \frac{(1+iy)^{-3/2}}{e^{2\pi y}-1}\,dy=5.224751\cdots$$ I don't think this integral has a closed-form expression.

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Thanks to Carlo, using Abel--Plana once again one finds the simple formula $$\lim_{n\to\infty}n^{3/2}S_n=2\zeta(3/2)$$

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  • $\begingroup$ wow, so this integral does have a closed form expression, cool! $\endgroup$ Commented Apr 8 at 18:36
  • $\begingroup$ @CarloBeenakker: well, whether $\zeta(3/2)$ is a "closed form expression" is debatable, no? $\endgroup$ Commented Apr 9 at 1:01

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