Show $\lim\limits_{n \to \infty } {a_n} = \frac{1}{\pi }{\operatorname{B}}\left( {\frac{1}{2},\frac{3}{4}} \right)$

Question

Let $a_1= 0,a_2=1$ and $a_n = \sqrt {\frac{{{a_{n - 1}} + {a_{n - 2}}}}{2} \cdot {a_{n - 1}}}$ for $n \geqslant 3$. Is it true that $\lim\limits_{n \to \infty } {a_n} = \frac{1}{\pi}{\operatorname{B}}\left( {\frac{1}{2},\frac{3}{4}} \right)$, where $\operatorname{B}(x_1,x_2)$ is the Beta function?

Having experimented with Maple and wxMaxima, I think this statement should be provable(?). For example, $a_{20}=.7627597635$ while $\frac{1}{\pi }{\rm B}\left( {\frac{1}{2},\frac{3}{4}} \right)=.7627597633...$.

I was also wondering whether this sequence is convergent, and I think I can show that. In fact, the subsequence $\langle a_{2n+1} \mid n\in\mathbb{N} \rangle$ is increasing while $\langle a_{2n} \mid n\in\mathbb{N} \rangle$ is decreasing and both subsequences are bounded, so they converge. Moreover, their limits are the same, so $\langle a_{n} \mid n\in\mathbb{N} \rangle$ converges, and even rough estimates show that the limit lies in $(0.758,0.766)$. But to show that it is exactly $\frac{1}{\pi }{\rm B}\left( {\frac{1}{2},\frac{3}{4}} \right)$ is more difficult. Nonetheless, I will write down the proof of convergence. Perhaps, it may hint somehow to ${\operatorname{B}}\left( {\frac{1}{2},\frac{3}{4}} \right)$ expressed as an infinite sum or infinite product or whatever.

---

Let's start with the observation that for all $n$: $0\leq a_{n}\leq1$. This holds since given two values in the interval $[0,1]$, their average and product lie again in this interval. So in particular the sequence $\langle a_{n} \mid n\in\mathbb{N} \rangle$ is bounded. What’s more, the subsequence $\langle a_{2n+1} \mid n\in\mathbb{N} \rangle$ is increasing while $\langle a_{2n} \mid n\in\mathbb{N} \rangle$ is decreasing. We shall prove this by induction, making use of the well known inequality $\sqrt{x y}\leq{\frac{x+y}{2}}$. Hence we get $$\tag{1}a_{n-2}^{\frac{1}{4}}\cdot a_{n-1}^{\frac{3}{4}}\leq a_{n}\leq\frac{a_{n-2}}{4}+\frac{3a_{n-1}}{4}$$

Now we can prove that for all $n$ we have

$$\tag{2}a_{1}\lt \cdots\lt a_{2n-1}$$

$$\tag{3}a_{2}\gt \cdot\cdot\cdot\gt a_{2n}$$

$$\tag{4}a_{2n-1}\lt a_{2n}$$

For $n=1$ we see that (2) and (3) are vacuously true and (4) is obvious. And for the induction step we have to show $$\tag{5}a_{2n-1}\lt a_{2n+1}\lt a_{2n+2}\lt a_{2n}$$

We start by observing that $$\tag{6}a_{2n+1}\lt a_{2n}$$

since by definition $\displaystyle a_{2n+1}=\sqrt{\frac{a_{2n-1}+a_{2n}}{2}\cdot a_{2n}}$ and from (4) it follows that $\displaystyle{\frac{a_{2n-1}+a_{2n}}{2}}\,\lt \,a_{2n}$.

Now for the first inequality of (5), note that by the first inequality of (1) we have $a_{2n+1}^{4}\ge a_{2n-1}\cdot a_{2n}^{3}$, and by (6) this is greater than $a_{2n-1}\cdot a_{2n+1}^{3}$, so $a_{2n+1}\gt a_{2n-1}$ as required. Similarly, $a_{2n+2}^{4}\geq a_{2n}\cdot a_{2n+1}^{3}\gt a_{2n+1}^{4}$, which proves the second inequality of (5). For the last inequality of (5) we use the second inequality of (1), so $\displaystyle a_{2n+2}\,\leq\,\frac{a_{2n}+3a_{3n+1}}{4\,a n+1}\,\lt \,a_{2n}$ (by (6)). That completes the induction proof.

---

Now we can easily prove that both subsequences converge to the same limit. This is because by (1) $\displaystyle a_{2n+2}-a_{2n+1}\leq{\frac{a_{2n}-a_{2n+1}}{4}}$ and by (5) this last difference is less than $\displaystyle\frac{a_{2n}-a_{2n-1}}{4}$. So the distance between $a_{2n+2}$ and $a_{2n+1}$ is less than a quarter of the distance between $a_{2n}$ and $a_{2n-1}$, and therefore these distances go to zero. That implies both subsequences have the same limit.

Regarding the value $l$ of the limit, if - inspired by (1) - we define the sequence $\langle b_{n}|n\in\mathbb{N}\rangle $ by $\displaystyle b_{n}={\frac{b_{n-2}}{4}}+{\frac{3b_{n-1}}{4}}$ and $b_1=0,b_2=1$, then clearly $a_{n}\leq b_{n}$. Also, it is not hard to prove that $\displaystyle b_{n}=\frac{4}{5}\left(1-\left(-{\frac{1}{4}}\right)^{n-1}\right)$, which converges to $0.8$. In the same way, if we define $c_{n}=c_{n-2}^{\frac{1}{4}}\cdot c_{n-1}^{\frac{3}{4}}$, then we better ignore the initial value $0$, so we just put $c_{2}=1,c_{3}=a_{3}={\sqrt{\frac{1}{2}}}$ and when we switch to $d_{n}:=\log_{2}c_{n}$, then $d_{2}=0,d_{3}=-{\frac{1}{2}}$ and $d_n$ satisfies the same recurrence relation as $b_n$. Therefore, for $n≥2$, $\displaystyle d_{n}=\frac{2}{5}\left(-1+\left(-\frac{1}{4}\right)^{n-2}\right)$, which converges to $-\frac{2}{5}$, so $c_n$ converges to $2^{-{\frac{2}{5}}}$, which is approximately $0.758$.

In fact, returning to $b_n$, but using initial values $b_{2}=1,b_{3}=a_{3}={\sqrt{\frac{1}{2}}}$, we can show $b_n$ converges to $\displaystyle\frac{1+2{\sqrt{2}}}{5}$, which is approximately $0.766$.

Answer 1

First note that, if $a_0=0,a_1=1$ and $a_{n+2}=\sqrt{a_{n+1}\left(\frac{a_n+a_{n+1}}{2}\right)}$, we can define $(x_n)_{n\geq0}$ as $x_n=\frac{a_n}{a_{n+1}}$ with initial value $x_0=0$ and the recursion formula $$x_{n+1}=\frac{a_{n+1}}{a_{n+2}}=\frac{a_{n+1}}{\sqrt{a_{n+1}\left(\frac{a_n+a_{n+1}}{2}\right)}}=\sqrt{\frac{2}{1+\frac{a_n}{a_{n+1}}}}=\sqrt{\frac{2}{1+x_n}}$$ So $a_n=\prod\limits_{k=1}^nx_k^{-1}$ and $a_\infty:=\lim_n a_n=\prod\limits_{k=1}^\infty x_k^{-1}$. Now consider the Weierstrass elliptic function $\wp_1(z):=\wp(z,1,0)$ satisfying $\wp_1'(z)^2=4\wp_1(z)^3-\wp_1(z)$ and define $$\mathfrak{P}(z):=\frac{\wp_1(z)^2+\frac{1}{4}}{\wp_1(z)^2-\frac{1}{4}}\ \text{ with }\ \mathfrak{P}(0)=\lim_{z\to0}\mathfrak{P}(z)=\lim_{z\to0}\frac{1+\frac{1}{4}\wp_1(z)^{-2}}{1-\frac{1}{4}\wp_1(z)^{-2}}=1$$

$\color{#FF5200}{\bf{Lemma\ 1.}}$ The function $\mathfrak{P}$ satisfies the two following identities $$\mathfrak{P}\left(\frac{z}{1+i}\right)=\sqrt{\frac{2}{1+\mathfrak{P}(z)}},\quad \mathfrak{P}\left((1+i)z\right)=\frac{2}{\mathfrak{P}(z)^2}-1$$

Proof. First notice that $\wp_1(iz)=\wp(iz,1,0)=-\wp(z,1,-0)=-\wp_1(z)$ and so $\wp_1'(iz)=\frac{1}{i}\frac{d}{dz}\left[\wp_1(iz)\right]=i\wp_1'(z)$. By the addition formula of Weierstrass elliptic functions, $$\wp_1((1+i)z) = \frac{1}{4}\left[\frac{\wp_1'(z)-\wp_1(iz)}{\wp_1(z)-\wp_1(iz)}\right]^2-\wp_1(z)-\wp_1(iz)$$ $$= \frac{1}{4}\left[\frac{\wp_1'(z)-i\wp_1(z)}{\wp_1(z)+\wp_1(z)}\right]^2-\wp_1(z)+\wp_1(z)=\frac{1}{4}\left[\frac{(1-i)\wp_1'(z)}{2\wp_1(z)}\right]^2=\frac{1}{8i}\frac{\wp_1'(z)^2}{\wp_1(z)^2}$$ $$=\frac{1}{8i}\frac{4\wp_1(z)^3-\wp_1(z)}{\wp_1(z)^2}=\frac{1}{2i}\left(\wp_1(z)-\frac{1}{4}\wp_1(z)^{-1}\right)$$ Now we can prove the second expression as $$\mathfrak{P}\left((1+i)z\right)=\frac{\wp_1((1+i)z)^2+\frac{1}{4}}{\wp_1((1+i)z)^2-\frac{1}{4}}=\frac{\frac{-1}{4}\left(\wp_1(z)^2-\frac{1}{2}+\frac{1}{16}\wp_1(z)^{-2}\right)+\frac{1}{4}}{\frac{-1}{4}\left(\wp_1(z)^2-\frac{1}{2}+\frac{1}{16}\wp_1(z)^{-2}\right)-\frac{1}{4}}$$ $$\frac{\wp_1(z)^2-\frac{1}{2}+\frac{1}{16}\wp_1(z)^{-2}-1}{\wp_1(z)^2-\frac{1}{2}+\frac{1}{16}\wp_1(z)^{-2}+1}=\frac{\wp_1(z)^2-\frac{3}{2}+\frac{1}{16}\wp_1(z)^{-2}}{\wp_1(z)^2+\frac{1}{2}+\frac{1}{16}\wp_1(z)^{-2}}$$ $$=2\frac{\wp_1(z)^2-\frac{1}{2}+\frac{1}{16}\wp_1(z)^{-2}}{\wp_1(z)^2+\frac{1}{2}+\frac{1}{16}\wp_1(z)^{-2}}-1=2\left(\frac{\wp_1(z)-\frac{1}{4}\wp_1(z)^{-1}}{\wp_1(z)+\frac{1}{4}\wp_1(z)^{-1}}\right)^2-1 = \frac{2}{\mathfrak{P}(z)^2}-1$$ and the first one follows from the second. $\square$

The former expression is crucial as, if we find $z_0\in\mathbb{C}$ such that $\mathfrak{P}(z_0)=0$, it let's us express $$\forall k\in\mathbb{N}:x_k=\mathfrak{P}\left(\frac{z_0}{(1+i)^k}\right)\Rightarrow a_\infty=\prod_{k=1}^\infty\mathfrak{P}\left(\frac{z_0}{(1+i)^k}\right)^{-1}$$ so our main goal is to obtain the root $z_0$ and, then, compute the infinite product.

$\color{#FF5200}{\bf{Lemma\ 2.}}$ The equation $\mathfrak{P}(z_0)=0$ has the solution $z_0=\frac{(-1)^{-3/4}}{4}B\left(\frac{1}{2},\frac{1}{4}\right)$

Proof. Consider $\mathfrak{P}(z_0)=0$. First note that, differentiating, $$\mathfrak{P}'(z)=-\frac{\wp_1(z)\wp_1'(z)}{\left(\wp_1(z)^2-\frac{1}{4}\right)^2}=-\frac{\wp_1(z)\sqrt{4\wp_1(z)\left(\wp_1^2(z)-\frac{1}{4}\right)}}{\left(\wp_1(z)^2-\frac{1}{4}\right)^2}$$ $$ = -\frac{2\wp_1(z)^{3/2}}{\left(\wp_1(z)^2-\frac{1}{4}\right)^{3/2}}=-2\left(\frac{\wp_1(z)^2}{\left(\wp_1(z)-\frac{1}{4}\right)^2}\right)^{3/4}$$ $$=-2\left(\frac{\left(\wp_1(z)+\frac{1}{4}\right)^2}{\left(\wp_1(z)-\frac{1}{4}\right)^2}-1\right)^{3/4}=-2(\mathfrak{P}(z)^2-1)^{3/4}$$ And, as $\mathfrak{P}(0)=1$ and $\mathfrak{P}(z_0)=0$, we have $$z_0=z_0-0=\mathfrak{P}^{-1}(0)-\mathfrak{P}^{-1}(1)=-\int_0^1(\mathfrak{P}^{-1})'(t)dt=-\int_0^1\frac{1}{\mathfrak{P}'(\mathfrak{P}^{-1}(t))}dt$$ $$=\frac{1}{2}\int_0^1\frac{1}{(t^2-1)^{3/4}}dt=\frac{(-1)^{-3/4}}{2}\int_0^1(1-t^2)^{-3/4}dt$$ and, substituting $t=\sqrt{s}\Rightarrow dt=\frac{1}{2}s^{-1/2}ds$, we finally see that $$z_0=\frac{(-1)^{-3/4}}{2}\int_0^1(1-s)^{-3/4}\frac{1}{2}s^{-1/2}ds=\frac{(-1)^{-3/4}}{4}B\left(\frac{1}{2},\frac{1}{4}\right)\ \square$$

$\color{#FF5200}{\bf{Lemma\ 3.}}$ The following infinite product holds $$\mathfrak{P}'(z)=2z^3\prod_{k=1}^\infty\mathfrak{P}\left(\frac{z}{(1+i)^k}\right)^{-3}$$

Proof. First observe that, via recursion, we have $$\mathfrak{P}'(z)=\frac{d}{dz}\left[\mathfrak{P}(z)\right]=\frac{d}{dz}\left[\frac{2}{\mathfrak{P}\left(\frac{z}{1+i}\right)^2}-1\right]=-4\frac{\mathfrak{P}'\left(\frac{z}{1+i}\right)}{1+i}\left(\mathfrak{P}\left(\frac{z}{1+i}\right)^{-3}\right)$$ $$=(-4)^n\frac{\mathfrak{P}'\left(\frac{z}{(1+i)^n}\right)}{(1+i)^n}\prod_{k=1}^n \mathfrak{P}\left(\frac{z}{(1+i)^k}\right)^{-3} =(1+i)^{3n}\mathfrak{P}'\left(\frac{z}{(1+i)^n}\right)\prod_{k=1}^n \mathfrak{P}\left(\frac{z}{(1+i)^k}\right)^{-3}$$ where we have used Lemma 1 and the fact that $-4=(1+i)^4$. Now, it is not hard to prove that $\mathfrak{P}(z)=1+\frac{1}{2}z^4+...$ so $\mathfrak{P}'(z)=2z^3+...$ and $$\lim_n(1+i)^{3n}\mathfrak{P}'\left(\frac{z}{(1+i)^n}\right)=2z^3\lim_n\frac{\mathfrak{P}'\left(\frac{z}{(1+i)^n}\right)}{2\left(\frac{z}{(1+i)^n}\right)^3}=2z^3\lim_{s\to0}\frac{\mathfrak{P}'(s)}{2s^3}=2z^3$$ Hence, taking $n\to+\infty$, we finally obtain $$\mathfrak{P}'(z)=\lim_n\left[(1+i)^{3n}\mathfrak{P}'\left(\frac{z}{(1+i)^n}\right)\right]\prod_{k=1}^\infty\mathfrak{P}\left(\frac{z}{(1+i)^k}\right)^{-3}=2z^3\prod_{k=1}^\infty\mathfrak{P}\left(\frac{z}{(1+i)^k}\right)^{-3}\ \square$$

$\color{#FC3503}{\bf{Theorem.}}$ We can express the limit $a_\infty=\lim_na_n$ as $$a_\infty=\prod_{k=1}^\infty\mathfrak{P}\left(\frac{(-1)^{-3/4}B\left(\frac{1}{2},\frac{1}{4}\right)}{4(1+i)^{k+1}}\right)^{-1}=\frac{1}{\pi}B\left(\frac{1}{2},\frac{3}{4}\right)$$

Proof. Isolating the infinite product from the previous expression, we see $$a_\infty=\prod_{k=1}^\infty\mathfrak{P}\left(\frac{z_0}{(1+i)^{k+1}}\right)^{-1}=\frac{1}{z_0}\sqrt[3]{\frac{\mathfrak{P}'(z_0)}{2}}=-\frac{4(-1)^{3/4}}{B\left(\frac{1}{2},\frac{1}{4}\right)}\sqrt[3]{(\mathfrak{P}(z_0)^2-1)^{3/4}}$$ $$=-\frac{4(-1)^{3/4}}{B\left(\frac{1}{2},\frac{1}{4}\right)}\sqrt[3]{(-1)^{3/4}}=-\frac{4(-1)^{3/4}(-1)^{1/4}}{B\left(\frac{1}{2},\frac{1}{4}\right)}=\frac{4}{B\left(\frac{1}{2},\frac{1}{4}\right)}$$ where $z_0=\frac{(-1)^{-3/4}}{4}B\left(\frac{1}{2},\frac{3}{4}\right)$. So, using the value $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$, we conclude that $$a_\infty=\frac{4\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{4}\right)}=\frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{5}{4}\right)}=\frac{1}{\pi}\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{5}{4}\right)}=\frac{1}{\pi}B\left(\frac{1}{2},\frac{3}{4}\right)\ \square$$