We know solution to infinite nested radicals such as $\sqrt{2+\sqrt{2+\sqrt{2+...}}} = 2$
But when we subtract this infinite nested radical from 2 and again sqrt it, it is not zero.
To quote example, in one form of Viète formula for $$\pi = \lim_{n\to\infty} 2^n\cdot\sqrt{2-\sqrt{2+\sqrt{2+... (\text{n - 1 times})}}}$$ https://en.m.wikipedia.org/wiki/Viète%27s_formula
Geometrically this is square inscribed in in unit circle, with many iterations of doubling the number of polygons. $2^n$ represents the number of polygons and infinite nested square roots of 2 represents side of the single polygon which is very small.
My question is when we have infinite nested radicals with such solutions when we do some extra operation(s) over that(like subtraction), does the result always change? If it changes what is the proof of theorem?
It means $a_{n+1}=\sqrt{2+a_n},$ where $a_1=\sqrt2$ and now we can calculate $\lim\limits_{n\rightarrow+\infty}a_n,$ which is definitely.