# What's equal the below power nested radical?

The same copy of this question is montioned here in SE with no convinced Answer , I want to know what MO will say about the below nested radical as a power form

it is well known that

$$\frac{2}{\pi}=\sqrt{\frac12}{\sqrt{\frac12+\frac12\sqrt{\frac12}}{\sqrt{\frac12+\frac12\sqrt{\frac12+\frac12\sqrt{\frac12}}}{\sqrt{\frac12+\frac12\sqrt{\frac12\cdots}}}}}$$

My Idea is to know what about above product if it is a power as shown below : $$A=\sqrt{\frac12}^{\sqrt{\frac12+\frac12\sqrt{\frac12}}^{\sqrt{\frac12+\frac12\sqrt{\frac12+\frac12\sqrt{\frac12}}}^{\sqrt{\frac12+\frac12\sqrt{\frac12\cdots}}}}}$$ ?

• This is the same question. You define (at least I think you do, you didn't indicate in what order exactly the exponential is supposed to be unwrapped, but I'll read it from left to right) $a_1=1/\sqrt{2}$, and then $a_n=a_{n-1}^{q_n}$, with $q_n$ denoting the $n$th square root in the original expression. So $\log a_n = q_n \log a_{n-1} = q_n \cdots q_2\log a_1$, and now you get your answer from the original result. – Christian Remling Dec 27 '18 at 23:02