If $\displaystyle a_0=\frac12$ and $\displaystyle a_{n+1}=\frac{1-\sqrt{1-a_n}}{1+\sqrt{1-2a_n}}$, show that
$$\sum_{n=0}^\infty2^na_{n+1}\sqrt{a_n}=\frac{\Gamma^2(\frac14)-4\cdot\Gamma^2(\frac34)}{8\sqrt{2\pi}}$$
The closed form for this series involves the square of gamma function, hence I try to connect it with integrals, which requires us first to find the explicit form from the recursion equation. But this recusion equation is highly non-linear. I try to multiply $1-\sqrt{1-2a_n}$ to rationalize the denominator but seems no help. I also tried some non-linear sub, such as
$$\tan(x-y)=\frac{\tan x-\tan y}{1+\tan x\tan y}$$
where $x=\frac\pi4, \tan y=\sqrt{1-a_n}$, but the $2a_n$ term inside the square root kills this attempt. If let $a_n=\sin^2\theta_n$, then we get
$$1-\cos^2\theta_{n+1}=\sin^2\theta_{n+1}=\frac{1-\cos\theta_n}{1+\sqrt{\cos2\theta_n}}$$
Is there any hint? Thank you!
Best Answer
I got the basic idea for this post from Find the limit of $4^n a_n$, for the recurrent sequence $a_{n+1}=\frac{1-\sqrt{1-a_n}}{1+\sqrt{1+a_n}}$
The recurrence relation for $a_n$ can be more properly understood in terms of Jacobian elliptic function. I use the notation $\text{sn} (u, k) $ with $k$ as modulus. The Wikipedia entry uses $\text{sn} (u, m) $ with $m=k^2$ as parameter. We have by definition $$u=\int_0^{\text{sn}(u,k)}\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}\tag{1}$$ The function $\text{sn} (u, k) $ satisfies the half argument formula $$\text{sn} ^2(u/2,k)=\frac{1-\sqrt{1-\text{sn}^2(u,k)}}{1+\sqrt {1-k^2\text{sn}^2(u,k)} }\tag{2}$$ Putting $k^2=2$ and setting $a_n=\text{sn} ^2(u_n,\sqrt{2})$ we see that recursion for $a_n$ leads to $u_{n+1}=u_n/2$ via $(2)$ and hence $u_n=u_0/2^n$.
Since $a_0=1/2$ we have $\text{sn} (u_0,\sqrt{2})=1/\sqrt{2}$ and hence $$u_0=\int_0^{1/\sqrt{2}}\frac{dx}{\sqrt{(1-x^2)(1-2x^2)}}=\int_0^{\pi/4}(1-2\sin^2x)^{-1/2}\,dx$$ The integral above equals $$\frac{1}{2}\int_0^{\pi/2}(\cos x) ^{-1/2}\,dx=\frac{1}{4}B(1/4,1/2)=\frac{\Gamma^2(1/4)}{4\sqrt{2\pi}}$$ The series in question can now be rewritten as $$\sum_{n\geq 0}2^n\text{sn}(u_0/2^n)\text{sn}^2(u_0/2^{n+1})$$ and the expected sum equals $(u_0/2)-\pi/8u_0$.
The term $\pi/4u_0$ is actually the value of another integral $$\int_0^{\pi/4}(1-2\sin^2x)^{1/2}\,dx$$ which equals $B(3/4,1/2)/4=\Gamma^2(3/4)/\sqrt{2\pi}$ and hence we can observe that the expected sum of series is $$\frac{1}{2}\left(\int_0^{\pi/4}\frac{dx}{\sqrt{1-2\sin^2x}}-\int_{0}^{\pi/4}\sqrt{1-2\sin^2x}\,dx\right)=\int_0^{\pi/4}\frac{\sin^2x}{\sqrt{1-2\sin^2x}}\,dx$$
User K B Dave mentions in comments about the Jacobi Epsilon function $$\mathcal{E} (u, k) =\int_0^{\text{sn}(u,k)}\sqrt {\frac{1-k^2t^2}{1-t^2}}\,dt=\int_0^u\text{dn}^2(t,k)\,dt$$ which satisfies the relationship $$\mathcal{E} (u+v, k)=\mathcal{E}(u, k)+\mathcal{E}(v,k)-k^2\text {sn}(u, k) \text{sn} (v, k) \text{sn} (u+v, k) $$ Replacing $u, v$ both by $u/2$ and $k^2$ by $2$ we get $$2\operatorname{sn}^2(u/2)\operatorname{sn}u= 2\mathcal{E}(u/2)-\mathcal{E} (u) $$ Putting $u=u_0/2^n$ and multiplying the equation by $2^{n-1}$ we get $$2^n\operatorname{sn}^2(u_0/2^{n+1})\operatorname{sn}(u_0/2^n)=2^n\mathcal{E}(u_0/2^{n+1})-2^{n-1}\mathcal{E}(u_0/2^n)$$ and hence our series evaluates to $$\lim_{n\to\infty} 2^n\mathcal{E}(u_0/2^{n+1})-\frac{1}{2}\mathcal{E}(u_0)$$ The second term above is $$\frac{1}{2}\int_0^{\pi/4}\sqrt{1-2\sin^2t}\,dt$$ and it remains to prove that $2^n\mathcal{E}(u_0/2^{n+1})\to u_0/2$ which equivalently requires us to prove that $\mathcal{E} (u) /u\to 1$ as $u\to 0$. This is luckily an easy consequence of fundamental theorem of calculus.
I should have remembered this Jacobian Epsilon function because I had asked about a proof of its addition formula some years ago. Damn!!