Astonishing: the sum of two infinite products of nested radicals equal to $ \pi $. Recently, I found the following handwritten expression in an old math book of my family. Probably it belonged to my great grandfather Boris, who had a P.D. in mathematics.
$ \pi = \frac{4\sqrt{5}}{5}.\frac{2}{\sqrt{2 + \frac{4}{\sqrt{5}}}}.\frac{2}{\sqrt{2 + \sqrt{2 +\frac{4}{\sqrt{5}}}}}.\frac{2}{\sqrt{2 + \sqrt{2 + \sqrt{2 +\frac{4}{\sqrt{5}}}}}}... +\frac{2\sqrt{10}}{5}.\frac{2}{\sqrt{2 + \frac{6}{\sqrt{10}}}}.\frac{2}{\sqrt{2 + \sqrt{2 +\frac{6}{\sqrt{10}}}}}.\frac{2}{\sqrt{2 + \sqrt{2 + \sqrt{2 +\frac{6}{\sqrt{10}}}}}}... $
I found this identity extremely interesting, and, in fact, I had never seen it. It's similar, but different, from Vieta's formula for $ \pi $
How to prove this identity?

Let $\theta=\arctan\frac{1}{2}$. Then $\cos\theta = \frac{2}{\sqrt{5}}$, hence $\sec\theta=\frac{\sqrt{5}}{2}$ and
$$\cos\frac{\theta}{2}=\sqrt{\frac{1}{2}+\frac{1}{\sqrt{5}}},\qquad \cos\frac{\theta}{4}=\sqrt{\frac{1+\sqrt{\frac{1}{2}+\frac{1}{\sqrt{5}}}}{2}}$$
and so on. Perform a bit of maquillage, then recall that
$$ \cos(\theta)\cos(2\theta)\cdots\cos(2^N\theta)=\frac{\sin(2^{N+1}\theta)}{2^N\sin(\theta)}$$
by a telescopic product, and you will recognize that your identity is just asserting
$$ \arctan\frac{1}{2}+\arctan\frac{1}{3}=\frac{\pi}{4},$$
pretty well-known. Your grandfather just combined a Machin-like formula with the principle behind Vieta's formula for $\pi$.
