Problem :
Show that if :
$$a=\int_{0}^{1}x!dx,b=\int_{0}^{\infty}1/\Gamma(x)dx,x!=\Gamma(x+1)$$
Then we have :
$$S=\frac{\pi^2}{6}\sqrt{ab+\sqrt{ab+2\sqrt{ab+3\sqrt{\cdots}}}}=4.0054\cdots>4$$
There many things related to the Basel constant as the probability to get a prime number for example .
The Fransen Robinson constant have nice integral represntation with exponential notably .
Note that the problem is similar to nested radicals due to Ramanujan .
I hope to see nice proof which could use simple bound or Cauchy-Schwarz inequality .
Dig the problem :
Using The Infinitely Nested Radicals Problem and Ramanujan's wondrous formula :
Cheating one radicals we have for :
$$2(2u+v)=(u+v)^2-ab$$
We get $u\simeq -1.795,v\simeq 2.795$
Introducing :
$$f(x)=ux+v$$
Setting :
$$uX+v=0$$
We have :
$$ |S6/\pi^2-ab|\simeq X/10$$
Already it can be done by hand if we have a good approximation for the integrals .
A relatively good approximation is :
$$\int_{0}^{1}1/576(-x^5+x^3+6x^2-6x+24)^2-x!dx>0$$
Nota bene :
As in my comment we can reach an equality for some constant $m,x$ :
$$\sqrt{ab+\sqrt{ab+2\sqrt{ab+3\sqrt{\cdots}}}}\left(1+1/2^2+1/3^2+\cdots+1/m^2+1/(m+1-x)^2+1/(m+1+x)^2\right)=4$$
For a bit of geometry see also Fuss's theorem for bicentric quadrilateral .
Question :
How to show it without the help of any numerical assistance so by hand ?
It works
I used $a$ and $b$ with $500$ decimal places.
As a function of $n$, the results are $$\left( \begin{array}{cc} 1 & 3.3713015211722886250 \\ 2 & 3.6786518328632658644 \\ 3 & 3.8317135361506168988 \\ 4 & 3.9125545019488574270 \\ 5 & 3.9559753173774284169 \\ 6 & 3.9792859467378086387 \\ 7 & 3.9917168350426439386 \\ 8 & 3.9982924950986951448 \\ 9 & 4.0017446141959247234 \\ 10 & 4.0035452069065250960 \\ 20 & 4.0054686797256613817 \\ 30 & 4.0054709810658212318 \\ 40 & 4.0054709835714249571 \\ 50 & 4.0054709835740579318 \\ 60 & 4.0054709835740606477 \\ 70 & 4.0054709835740606505 \\ 80 & 4.0054709835740606505 \\ \end{array} \right)$$
For $n=167$, fifty exact significant figures.