If $\displaystyle a_0=\frac12$ and $\displaystyle a_{n+1}=\frac{1-\sqrt{1-a_n}}{1+\sqrt{1-2a_n}}$, show that


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


Is there any hint? Thank you!

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    $\begingroup$ Can you tell us where the problem comes from? $\endgroup$
    – Martin R
    Commented Jun 19, 2023 at 18:47
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    $\begingroup$ Found a better post to link I think: math.stackexchange.com/a/229265/1104384 $\endgroup$
    – Bruno B
    Commented Jun 19, 2023 at 19:37
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    $\begingroup$ I add that if someone manages to prove that the sum is equivalent to $$\frac{\sqrt{\pi}}{4}\sum_{n=0}^\infty(-1)^n \frac{\Gamma\left(\frac{2n+3}{4}\right)}{\Gamma\left(\frac{2n+5}{4}\right)}$$ then the proof is complete, just by using this result: math.stackexchange.com/q/4717394/1174256 $\endgroup$
    – Zima
    Commented Jun 19, 2023 at 20:28
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    $\begingroup$ This is false, since all terms of $a_n$ in my OP are postive, but your referenced series is alternating which includes negative terms. @Zima $\endgroup$
    – MathFail
    Commented Jun 19, 2023 at 21:32
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    $\begingroup$ Equality means nothing, because you can always make two convergent series equal to each other, for example, suppose two series converge to $a$ and $b$, then $a=\frac ab b$, but it doesn't mean they have the same structure. Here the OP has all positive terms, your referenced one are alternating, they have different structure, and this difference is intrinsic @Zima $\endgroup$
    – MathFail
    Commented Jun 19, 2023 at 21:35

2 Answers 2


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!!

  • $\begingroup$ It is not a telescoping series. Or maybe there are other ways to telescope it. $\endgroup$
    – MathFail
    Commented Jun 21, 2023 at 6:00
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    $\begingroup$ $2 \mathcal{E}(t/2) - \mathcal{E}(t)=k^2\,\mathrm{sn}^2\tfrac{t}{2}\,\mathrm{sn}\,t$, with $\mathcal{E}$ the Jacobi epsilon function, so... $\endgroup$
    – K B Dave
    Commented Jul 8, 2023 at 22:48
  • $\begingroup$ Thanks a lot @KBDave. Sometimes the solution is right there in front of our eyes but is not visible unless one gets some guidance. $\endgroup$
    – Paramanand Singh
    Commented Jul 9, 2023 at 2:06
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    $\begingroup$ A remark: this question doesn't rely on the extra symmetry of lemniscatic $k^2=2$, save for the reductions of the complete integrals to gamma products. An example that does: if $x_0=1$, $x_{2n+1}=2(x_{2n}+\sqrt{x_{2n}^2-1})$, $x_{2n+2}=\tfrac{1}{2}(x_{2n+1}+\sqrt{x_{2n+1}^2+4})$, then $\sum_{n=0}^{\infty}\tfrac{(-1)^n}{4^{\lfloor n/2\rfloor}x_n}=\tfrac{32\pi}{\Gamma(\tfrac{1}{4})^4}$. $\endgroup$
    – K B Dave
    Commented Jul 9, 2023 at 3:31
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    $\begingroup$ @KBDave: that calls for an interesting separate question. If time permits, maybe you can ask it as a self answered question with all relevant context. $\endgroup$
    – Paramanand Singh
    Commented Jul 9, 2023 at 4:33

Check @ParamanandSingh's initial conjecture, and show that it is NOT a telescoping series. Use this Reference Convension (RC) for Jacobi elliptic function. Let $a_n=\text{sn}^2(u_n,k=\sqrt2)$, for short, we suppress the index $k$ and denote $\text{sn}(u_n)=\text{sn}(u_n,k=\sqrt2)$, the recursion equation becomes:


Use the property (23), (24) in RC, we get


Use (69) in RC, we get

$$\text{sn}^2(u_{n+1})=\text{sn}^2(\frac{u_{n}}2)\Longrightarrow u_n=\frac{1}{2}u_{n-1}\Rightarrow \boxed{a_n=\text{sn}^2(\frac{u_{0}}{2^n})}$$

where $$u_0=\int_0^{\pi/4}\frac1{\sqrt{1-2\sin^2x}}dx=\int_0^{\pi/4}\cos^{-\frac12}(2x)dx=\frac14B\left(\frac12,\frac14\right)=\frac1{4\sqrt{2\pi}}\Gamma^2(\frac14)$$

the series becomes

$$S=\sum_{n=0} 2^n \text{sn}^2(\frac{u_n}2)\text{sn}(u_{n})$$

Use (69)

$$S=\sum_{n=0} 2^n\frac{1-\text{cn}(u_n)}{1+\text{dn}(u_n)} \text{sn}(u_{n})=\sum_{n=0} \frac{2^n\text{sn}(u_{n})}{1+\text{dn}(u_n)}-\frac{2^{n-1}\cdot2\text{sn}(u_n)\text{cn}(u_{n})}{1+\text{dn}(u_n)}\tag{*}$$

Use (63)


and (65)


Plug into (*), we get

$$S=\sum_{n=0} \frac{2^n\text{sn}(u_{n})}{1+\text{dn}(u_n)}-\frac{2^{n-1}\cdot\text{sn}(u_{n-1})}{1+\text{dn}(u_{n-1})}\cdot\frac{2(1-2\text{sn}^2(u_{n}))}{\text{dn}(u_n)(1+\text{dn}(u_n))}$$

Use (24)

$$S=\sum_{n=0} \frac{2^n\text{sn}(u_{n})}{1+\text{dn}(u_n)}-\frac{2^{n-1}\cdot\text{sn}(u_{n-1})}{1+\text{dn}(u_{n-1})}\cdot\color{red}{\frac{2\text{dn}(u_{n})}{1+\text{dn}(u_n)}}\tag{**}$$

Numerically checked, (*)$=$(**).

For $n=1, n=2, n=3$, the first half of (**) are

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\sum_{n=1}^3 \frac{2^n\text{sn}(u_{n})}{1+\text{dn}(u_n)}=0.658552 + 0.655704 + 0.655526$$

the second half of (**) are

$$\sum_{n=1}^3 \frac{2^{n-1}\cdot\text{sn}(u_{n-1})}{1+\text{dn}(u_{n-1})}\cdot\color{red}{\frac{2\text{dn}(u_{n})}{1+\text{dn}(u_n)}}=0.553774 + 0.623159 + 0.646898$$

Unfortunately, it is NOT a telescoping series, due to the annoying red term.

  • $\begingroup$ The expected sum is not the first term of the series, whence there is no reason to expect a telescoping property. $\endgroup$
    – Gary
    Commented Jun 21, 2023 at 6:18
  • $\begingroup$ If you read ParamanandSingh's comment (it is deleted now) and he first conjectured as a telescoping series, and I verified it is not. $\endgroup$
    – MathFail
    Commented Jun 21, 2023 at 6:21
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    $\begingroup$ I think there should be a recursion of the form $b_n=a_{n+1}\sqrt{a_n}+2b_{n+1}$ where $2^nb_n\to 0$. This gives the sum of series as $b_0$. My guess is that $b_n=\int_0^{\sqrt{a_n}}\frac{x^2}{\sqrt{(1-x^2)(1-x^2/a_n)}}\,dx$ but I don't have any way to verify this guess. It may be wrong but some sort of similar integral should work. $\endgroup$
    – Paramanand Singh
    Commented Jun 21, 2023 at 7:49
  • $\begingroup$ Yes, it means to find another way to telescope it, $2^na_{n+1}\sqrt{a_n}=2^nb_n-2^{n+1}b_{n+1}$. If this works, then the series $\sum_{n=0}2^na_{n+1}\sqrt{a_n}=b_0$ @ParamanandSingh $\endgroup$
    – MathFail
    Commented Jun 21, 2023 at 16:13
  • $\begingroup$ This kind of recursion is already used to evaluate certain elliptic integrals and hence I am hopeful here. See this blog post of mine paramanands.blogspot.com/2009/08/… (search for $L(a, b) $ there). $\endgroup$
    – Paramanand Singh
    Commented Jun 21, 2023 at 16:23

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