# Asymptotic behavior of the "Cauchy square" series

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

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 $$n \in \mathbb{N}$$, $$$$x_n \le c (\xi \alpha)^n n^{-\frac 32}. \qquad\qquad (EQ1)$$$$

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*}

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

## 2 Answers

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.

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

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