Closed form for $\left(1+\left(\frac{1}{2}+\left(\frac{1}{3}+\left(\frac{1}{4}+\cdots\right)^2\right)^2\right)^2\right)^2$?

Question

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.

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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.

Answer 1

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.