A simple and interesting recursion:
$$y_{n+1}=\frac{1}{2}(y_n+\sqrt{\frac{1}{2^{2n}}+y_n^2})$$
has these curious solutions
$$y_1=-\infty,y_{\infty}=\frac{1}{2\pi}$$ $$y_1=-\frac{1}{2},y_{\infty}=\frac{2}{3\pi}$$ $$y_1=0,y_{\infty}=\frac{1}{\pi} $$ $$y_1=\frac{1}{2},y_{\infty}=\frac{2}{\pi}$$
Cannot find it in the literature as such and it does not look like coming from AGM, but I suspect elliptic integrals. Still cannot start from anywhere for some time. Any ideas?
The closed form of the limit doesn't come from elliptic integral, but from half-angle formula for cotangent function. To illustrate this result, we will only condiser the case where $y_1 > 0$.
Consider the auxillary sequence $a_n = 2^n y_n$, it satisfies
$$a_{n+1} = a_n + \sqrt{1 + a_n^2}$$
If $a_n = \cot(\theta)$ for some $\theta \in (0,\frac{\pi}{2})$, then $$a_{n+1} = \frac{\cos\theta}{\sin\theta} + \sqrt{1 + \frac{\cos^2\theta}{\sin^2\theta}} = \frac{1+\cos\theta}{\sin\theta} = \cot\frac{\theta}{2}$$
Using this and the assumption $y_1 > 0$, we find following closed form expression of $y_{n+1}$.
$$2^{n+1}y_{n+1} = a_{n+1} = \cot\left(\frac{\cot^{-1}(2y_1)}{2^n}\right)$$
This leads to $$y_\infty = \lim_{n\to\infty} y_{n+1} = \lim_{n\to\infty} \frac{1}{2^{n+1}} \cot\left(\frac{\cot^{-1}(2y_1)}{2^n}\right) = \frac{1}{2\cot^{-1}(2y_1)}\quad\text{ whenever}\quad y_1 > 0$$