Show $\frac{\pi^2}{6}\sqrt{ab+\sqrt{ab+2\sqrt{ab+3\sqrt{\cdots}}}}>4$ by hand .

Question

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 ?

@ClaudeLeibovici Yeah nice problem found relatively quickly after a dream . If we use $\sum_{1}^{436} 1/n² =1.6426431$ and multiply by $2.4351$ we get a nice approximation or an almost integer .For your abacus (old computer?) you need some time and time after Stephen Hawking is numerics so we can try the old method pencil and paper ;-) . — Miss and Mister cassoulet char, Aug 22 '23 at 12:59
@ClaudeLeibovici Can you make an answer with numerical checking to enlight the problem ? — Miss and Mister cassoulet char, Aug 23 '23 at 12:13
Now, prove that, if $ab=2.577369193391029779713573708923493043415319293014$ the result is larger than $4$ (in fact, it is $4+2.773\times 10^{-49}$ — Claude Leibovici, Aug 29 '23 at 06:55

score 2 · Answer 1 · answered Aug 28 '23 at 15:04

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.

score 1 · Answer 2 · answered Sep 19 '23 at 12:46

1

We have $$ab = \frac {57}{22}-5.047\times 10^{-5} $$ and $S$ is an increasing function of $n$.

For $n=9$, $S=4.00177$.

answered Sep 19 '23 at 12:46

Claude Leibovici

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Show $\frac{\pi^2}{6}\sqrt{ab+\sqrt{ab+2\sqrt{ab+3\sqrt{\cdots}}}}>4$ by hand .

2 Answers2