Nested squares seem to be more promising than nested radicals, since they give rational approximations and in principle can be expanded into a series.
These two expressions converge numerically:
$$\left(1+\left(\frac{1}{2}+\left(\frac{1}{3}+\left(\frac{1}{4}+\cdots\right)^2\right)^2\right)^2\right)^2=2.14842827808221794391178636615$$
$$\left(1-\left(\frac{1}{2}-\left(\frac{1}{3}-\left(\frac{1}{4}-\cdots\right)^2\right)^2\right)^2\right)^2=0.680484597688804927729801584438$$
Search with ISC, Wolframalpha and OEIS did not reveal any closed forms for these numbers.
Is it possible that a closed form exists for these nested squares and how would you go about finding it?
The proper definition for the first nested square is the limit of the sequence:
$$s_1=1$$
$$s_2=\left(1+\left(\frac{1}{2}\right)^2\right)^2$$
$$s_3=\left(1+\left(\frac{1}{2}+\left(\frac{1}{3}\right)^2\right)^2\right)^2$$
Etc. The same for the second nested square.
Other two (alternating) expressions:
$$\left(1+\left(\frac{1}{2}-\left(\frac{1}{3}+\left(\frac{1}{4}-\cdots\right)^2\right)^2\right)^2\right)^2=1.27629973953623486796358849410$$
$$\left(1-\left(\frac{1}{2}+\left(\frac{1}{3}-\left(\frac{1}{4}+\cdots\right)^2\right)^2\right)^2\right)^2=0.462513422693928495067300679614$$
Again, I found nothing on these numbers.
If you know any reference about nested squares in general, it would be greatly appreciated as well.
Using the idea by TylerHG, we can approach the inner (infinity) part of expression to accelerate the convergence of the consequences. For the first sequence can be considered a function $$f(x) = \dfrac1{x+n}+f^2(x+1),\qquad(1)$$ or $$f(x-1) = \dfrac1{x+n-1} + f^2(x).$$ For $x\in[n-1,n]$ function $f(x)$ can be represented by Taylor series with sufficient accuracy, so $$f(x-1)\approx f(x)-f'(x)+\dfrac{f''(x)}2-\dfrac{f'''(x)}6+\dfrac{f^{(IV)}}{24}(x)-\dfrac{f^{(V)}(x)}{120}+\dfrac{f^{(VI)}(x)}{720}-\dots.$$ For arbitrary values of $x$ convenient to use the type of expansion $$f(x)=\dfrac1{x+n-1}\left(1+\dfrac{a_1}{x+n-1}+\dfrac{a_2}{(x+n-1)^2}+\dfrac{a_3}{(x+n-1)^3}+\dfrac{a_4}{(x+n-1)^4}+\dfrac{a_5}{(x+n-1)^5}+\dfrac{a_6}{(x+n-1)^6}+\dots\right),$$ then $$f^2(x) = \dfrac1{(x+n-1)^2}\left(1+\dfrac{2a_1}{x+n-1} + \dfrac{2a_2+a_1^2}{(x+n-1)^2} + \dfrac{2a_3+2a_1a_2}{(x+n-1)^3}\\+ \dfrac{2a_4+2a_1a_3+a_2^2}{(x+n-1)^4}+ \dfrac{2a_5+2a_1a_4+2a_2a_3}{(x+n-1)^5} + \dots\right),$$ $$-f'(x)=\dfrac1{(x+n-1)^2}\left(1+\dfrac{2a_1}{x+n-1}+\dfrac{3a_2}{(x+n-1)^2}+\dfrac{4a_3}{(x+n-1)^3}\\+\dfrac{5a_4}{(x+n-1)^4}+\dfrac{6a_5}{(x+n-1)^5}+\dots\right),$$ $$f''(x)=\dfrac1{(x+n-1)^3}\left(2+\dfrac{6a_1}{x+n-1}+\dfrac{12a_2}{(x+n-1)^2}+\dfrac{20a_3}{(x+n-1)^3}\\+\dfrac{30a_4}{(x+n-1)^4}+\dots\right),$$ $$-f'''(x)=\dfrac1{(x+n-1)^4}\left(6+\dfrac{24a_1}{x+n-1}+\dfrac{60a_2}{(x+n-1)^2}+\dfrac1{120}\dfrac{120a_3}{(x+n-1)^3}+\dots\right),$$ $$f^{IV}(x)=\dfrac1{(x+n-1)^5}\left(24+\dfrac{120a_1}{x+n-1}+\dfrac{360a_2}{(x+n-1)^2}+\dots\right),$$ $$-f^{V}(x)=\dfrac1{(x+n-1)^6}\left(120+\dfrac{720a_1}{x+n-1}+\dots\right),$$ $$f^{VI}(x)=\dfrac{720}{(x+n-1)^7}+\dots.$$ So we have: $$\begin{cases} a_1+1 = 1\\ a_2+2a_1+1 = 2a_1\\ a_3+3a_2+3a_1+1 = 2a_2+a_1^2\\ a_4+4a_3+6a_2+4a_1+1 = 2a_3+a_1a_2\\ a_5+5a_4+10a_3+10a_2+5a_1+1 = 2a_4+2a_1a_3+a_2^2\\ a_6+6a_5+15a_4+20a_3+15a_2+6a_1+1 = 2a_5 + 2a_1a_4+2a_2a_3,\\ a_7+7a_6+21a_5+35a_4+35a_5+21a_1+7a_1+1 = 2a_6 + 2a_1a_5 + 2a_2a_4+a_3^2, \end{cases}$$
from whence $$a_1=0;\quad a_2=-1,\quad a_3=0,\quad a_4=5,\quad a_5=-5,\quad a_6=-41,\quad a_7 = 145,$$ $$f(x)=\dfrac1{x+n-1} - \dfrac{1}{(x+n-1)^3}+\dfrac{5}{(x+n-1)^5}-\dfrac{5}{(x+n-1)^6}+\dfrac{-41}{(x+n-1)^7}+\dfrac{145}{(x+n-1)^8}+\dots.$$
Using the series obtained in the form of $$f(n)=\dfrac1n - \dfrac{1}{n^3}+\dfrac{5}{n^5}-\dfrac{5}{n^6}-\dfrac{41}{n^7}+\dfrac{145}{n^8}+\dots.$$ for calculating the supplement to the internal fractions (an infinite amount) significally increasing convergence of the original sequence.
Test results for $n = 8,$ gives $2.148428280400...,$ and this value approximates the limit $2.148428278082218...$ of first consequence with great precision.
When using the values n <8 resulting formula is a less accurate, but shorter.