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

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.

• I don't know if this helps, but the expression can be expressed as $\underset{n=1}{\overset{\infty}{\LARGE\mathrm F}} \ (\frac{1}{n}+x)^2$ – Jacob Apr 9 '16 at 2:15
• @Jacob What does the large F denote? – user236182 Apr 9 '16 at 2:23
• @user236182 This post should help you: math.stackexchange.com/questions/1224411/… – Jacob Apr 9 '16 at 3:14
• @Chip The Inverse Symbolic Calculator (ISC as mentioned by OP) is indeed one of the types of software you are describing. – Erick Wong Apr 24 '16 at 4:36
• Ahhh, that explains the confusion. OP means original post (or original poster if referring to the person rather than the content). ISC is Inverse Symbolic Calculator, and OEIS is Online Encyclopedia of Integer Sequences. – Erick Wong Apr 25 '16 at 6:54

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.

• Usually closed form means closed form ie expressing it as combinations of rationale and known irrationals.. Considering an approximation a "closed form" isn't really satisfactory. Of course, it can be useful nonetheless – Ant Jun 27 '16 at 15:52
• @Ant formulas for $f(8)$ and resulting formula are closed and contains only the 4 arithmetic operations. I broke them into two parts, so that the result could be seen in Wolfram Alpha – Yuri Negometyanov Jun 27 '16 at 15:57
• $f(8)$ is an infinite series, right? Not a closed form. – GEdgar Jun 27 '16 at 16:01
• @GEdgar, $f(8)$ was calculated as sum of seven fractions. Without it we have $2.147935291824$. – Yuri Negometyanov Jun 27 '16 at 16:03
• If it is not an exact expression for the original limiting value, it is not a 'closed form' for that value of the original series. – Steven Stadnicki Jun 27 '16 at 16:18