Show that : $\frac{1}{\zeta(3)}<2C-1$

Question

Problem :

Show that :

$$\frac{1}{\zeta(3)}<2C-1$$

Where we can see the zeta function and the Catalan's constant .

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It recall me the Faulhaber problem which relate the case $n=1,3$ with a square .

There are many representation for the Riemann's zeta function see https://functions.wolfram.com/ZetaFunctionsandPolylogarithms/Zeta/07/01/01/

Currently there is also nice representation for the Catalan's constant see https://en.wikipedia.org/wiki/Catalan%27s_constant

One hint :

For $x\geq 1$ we have :

$$\left(2C-1\right)\left(\sum_{n=1}^{100}\frac{x}{n^{3}}\right)^{-1}>\left(3-C\right)\frac{\ln\left(x\right)}{1+x^{2}}$$

But with all of this I cannot find a proof ...

Update :

Using positives function and Cauchy-Schwarz inequality we have :

$$\left(\int_{0}^{\infty}\left(\frac{x^{2}}{e^{x}-1}\frac{\left(\frac{x}{\cosh\left(x\right)}-\frac{1}{1.18293097883}x^{\frac{3}{2}}\cdot e^{-\frac{2}{3}x^{\frac{3}{2}}}\right)}{2}\right)^{\frac{1}{2}}dx\right)^{2}<\int_{0}^{\infty}\frac{\left(\frac{x}{\cosh\left(x\right)}-\frac{1}{1.18293097883}x^{\frac{3}{2}}\cdot e^{-\frac{2}{3}x^{\frac{3}{2}}}\right)}{2}dx\int_{0}^{\infty}\frac{x^{2}}{e^{x}-1}dx=1.000028\cdots$$

Now I really think we can use Cauchy-Schwarz for integral .

We have also empirically :

$$\frac{1}{\zeta(3)}<\frac{99486}{100000}\left(2\left(\int_{0}^{\infty}\frac{t}{\cosh\left(t+\frac{1}{t}\right)}dt\right)-1\right)\left(\int_{0}^{\infty}\frac{t}{e^{t}+1}dt\right)<2C-1$$

Update 2 :

Tricky Cauchy Schwarz inequality :

Let :

$$h\left(x\right)=\frac{2ax}{\cosh\left(ax\right)},g\left(x\right)=\frac{h\left(x\right)}{\int_{0}^{\infty}h\left(y\right)dy}$$

Then $\exists a ,1<a<\zeta(3)$ such that :

$$\frac{1}{a}\int_{0}^{\infty}\sqrt{\frac{x^{2}}{e^{x}+1}\cdot\frac{\left(h\left(x\right)-g\left(x\right)\right)}{2}}dx=1$$

Update 3 :

I finally found a reference see https://arxiv.org/pdf/2105.11771.pdf we have :

$$\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\frac{\ln\left(\cos\left(\frac{x}{2}\right)\right)-\ln\left(\cos\left(\frac{z}{2}\right)\right)}{\cos\left(x\right)-\cos\left(z\right)}dxdz=\pi G-\frac{7}{4}\zeta(3)$$

Then we can use $a,x\in[0,\pi/2)$ :

$$f\left(x\right)=\frac{\ln\left(\cos\left(\frac{x}{2}\right)\right)-\ln\left(\cos\left(\frac{a}{2}\right)\right)}{\cos\left(x\right)-\cos\left(a\right)},g\left(x\right)=\frac{e^{\frac{x}{4}}}{2}+\frac{e^{-\frac{x}{4}}}{2}-1+f\left(0\right),h\left(x\right)=f\left(x\right)-g\left(x\right)$$

Then we have :

$$f(x)\geq \frac{h''\left(0\right)}{2}x^{2}+\frac{h''''\left(0\right)}{4!}x^{4}+g\left(x\right)$$

Last update :

Why not invoke probability and Zeta function and probability and prime numbers https://arxiv.org/abs/2203.01832

How to show it without a calculator ? Is it an isloted problem and if so have you a reference ?

Ps: For the story I found it accidentaly in setting the value 0.831 in Wolfram Alpha see https://www.wolframalpha.com/input?i=0.831

Answer 1

Using, as @Calum Gilhooley suggested, the truncated series representation of the constants (these are generalized hypergeometric functions)

$$C_p=\sum_{k=1}^p\frac{4^{4 k-3} \left(580 k^2-184 k+15\right)}{k^3 (2 k-1) \binom{4 k}{2 k} \binom{6 k}{3 k} \binom{6 k}{4 k}}$$

$$\zeta_p(3)=\frac 1{64}\sum_{k=1}^p \frac{(-1)^k \left(205 k^2+250 k+77\right) (k!)^{10}}{((2 k+1)!)^5}$$ Define $$\Delta_p=\zeta_p(3)\,(2C_p-1)-1$$ $$\Delta_2=\frac{67988471}{29638224000000}=2.29395\times 10^{-6}$$ $$\Delta_=\frac{2311829612939580593}{81097164321132960000000}=2.85069\times 10^{-5}$$

Tedious (but doable) : using the "derivative" of the resulting generalized hypergeometric functions, we can show that $$\log(\Delta_{p+1}-\Delta_p)\sim 4.45 -5.22 \, p$$

Answer 2

Cutting Gordian knot by Mathematica:

N[1/Zeta[3]]

0.831907

N[2*Catalan - 1]

0.831931

Answer 3

I put here a failed attempt but if we find a refinement for Cauchy-Schwarz integral inequality perhaps it can be done :

let :

$$f\left(x\right)=\frac{x}{\cosh\left(x\right)}-\frac{x^{c^{\frac{1}{c^{d}}}}e^{-\frac{1}{c}x^{c}}}{\int_{0}^{\infty}t^{c^{\frac{1}{c^{d}}}}e^{-\frac{1}{c}t^{c}}dt},b=\frac{\left(\int_{0}^{\infty}f\left(ax\right)dx\right)}{\int_{0}^{\infty}f\left(x\right)dx},d=2,c=1.27,a=0.95$$

Then we have as numerical value :

$$1<1+\frac{\left(1-\left(\frac{\left(\int_{0}^{\infty}\left(\frac{\frac{x^{2}}{e^{x}-1}\left(f\left(ax\right)\right)}{2}\right)^{\frac{1}{2}}dx\right)^{2}}{b}\right)\right)}{6}<(\zeta(3))(2C-1)$$

In fact using Callebaut's inequality in a continuous form we have :

$\exists v\in[0.25,0.5)$ such that :

$$\int_{0}^{\infty}\left(\frac{x^{2}}{e^{x}-1}\right)^{-v+\frac{1}{2}}\left(\left(f\left(ax\right)\right)\right)^{v+\frac{1}{2}}dx\cdot\int_{0}^{\infty}\left(\frac{x^{2}}{e^{x}-1}\right)^{v+\frac{1}{2}}\left(f\left(ax\right)\right)^{-v+\frac{1}{2}}dx\cdot\frac{1}{2b}=1$$

Then we can separate the problem in two with $d,c,a$ as above and $v=0.48$ we have :

$$b^{\sqrt{\frac{77}{16}}}\int_{0}^{\infty}\left(\frac{x^{2}}{e^{x}-1}\right)^{-v+\frac{1}{2}}\left(\left(f\left(ax\right)\right)\right)^{v+\frac{1}{2}}dx>1$$$$\int_{0}^{\infty}\left(\frac{x^{2}}{e^{x}-1}\right)^{v+\frac{1}{2}}\left(f\left(ax\right)\right)^{-v+\frac{1}{2}}dx\cdot\frac{1}{2b^{\left(1+\sqrt{\frac{77}{16}}\right)}}>1$$

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Some numerical result we have :

$$\left(3\int_{0}^{\left(2-\sqrt{3}\right)}\frac{x-\frac{x^{3}}{3}+\frac{2}{15}x^{5}+7x\left(-\frac{x^{3}}{3}+\frac{2}{15}x^{5}\right)^{2}}{x}dx-\frac{1}{4}\pi\ln\left(2-\sqrt{3}\right)-1\right)\zeta(3)>1$$

$$\zeta(3)>1+\frac{1}{4+\frac{1}{1+\frac{1}{18+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{4+\frac{1}{1+\frac{1}{9+\frac{1}{9}}}}}}}}}}=177658/147795\tag{I}$$

Proof by hand of $I$

It's not hard to see as the number of fraction is even or odd then it's greater or less than the Apery's constant so we have :

$$\zeta(3)>1+\frac{1}{4+\frac{1}{1+\frac{1}{18+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{4+\frac{1}{1+\frac{1}{9+\frac{1}{9}}}}}}}}}}$$

Or :

$$1+\frac{1}{4+\frac{1}{1+\frac{1}{18+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{4+\frac{1}{1+\frac{1}{\frac{82}{9}}}}}}}}}}$$

Or :

$$1+\frac{1}{4+\frac{1}{1+\frac{1}{18+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{4+\frac{1}{\frac{91}{82}}}}}}}}}$$

Or :

$$1+\frac{1}{4+\frac{1}{1+\frac{1}{18+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{\frac{446}{91}}}}}}}}$$

Or :

$$1+\frac{1}{4+\frac{1}{1+\frac{1}{18+\frac{1}{1+\frac{1}{1+\frac{1}{\frac{537}{446}}}}}}}$$

Or :

$$1+\frac{1}{4+\frac{1}{1+\frac{1}{18+\frac{1}{1+\frac{1}{\frac{983}{537}}}}}}$$

Or :

$$1+\frac{1}{4+\frac{1}{1+\frac{1}{18+\frac{1}{\frac{1520}{983}}}}}$$

Or :

$$1+\frac{1}{4+\frac{1}{1+\frac{1}{\frac{28343}{1520}}}}$$

Or :

$$1+\frac{1}{4+\frac{1}{\frac{29863}{28343}}}$$

And at the end :

$$1+\frac{1}{\frac{147795}{29863}}=177658/147795$$

Doing the same matter for the Catalan's constant we have :

$$C>\frac{1}{1+\frac{1}{10+\frac{1}{1+\frac{1}{8+\frac{1}{1+\frac{1}{88}}}}}}=\frac{9690}{10579}$$

So we have :

$$(2\frac{9690}{10579}-1) 177658/147795=\frac{1563568058}{1563523305}>1$$

Answer 4

Too long for a comment :

If we show with Callebaut inequality :

$$\frac{9}{8}\int_{0}^{\frac{\pi}{2}}\left(\left(\left(-\frac{\ln\left(\tan\left(x\right)\right)}{4}+\ln\left(\sec\left(x\right)+\tan\left(x\right)\right)-\frac{x}{\int_{0}^{\frac{\pi}{2}}tdt}\right)\cdot\frac{4}{7}\left(x\ln\left(\sec\left(x\right)+\tan\left(x\right)\right)\right)\right)^{\frac{1}{2}}\right)dx<\int_{0}^{\frac{\pi}{2}}\frac{4}{7}\left(x\ln\left(\sec\left(x\right)+\tan\left(x\right)\right)\right)dx\int_{0}^{\frac{\pi}{2}}\left(\ln\left(\sec\left(x\right)+\tan\left(x\right)\right)-\frac{x}{\int_{0}^{\frac{\pi}{2}}tdt}\right)dx$$

Then one can show that :

$$\frac{9}{8}\int_{0}^{\frac{\pi}{2}}\left(\left(\left(-\frac{\ln\left(\tan\left(x\right)\right)}{4}+\ln\left(\sec\left(x\right)+\tan\left(x\right)\right)-\frac{x}{\int_{0}^{\frac{\pi}{2}}tdt}\right)\cdot\frac{4}{7}\left(x\ln\left(\sec\left(x\right)+\tan\left(x\right)\right)\right)\right)^{\frac{1}{2}}\right)dx>1$$

It seems promising to be continued ...

I think I'm not the first who find it so if someone have a reference...Anyway I set it in community wiki .

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Using Bernoulli's numbers , geometric series we have a rought approximation for $\zeta(3)$ we have :

$$\zeta(3)\simeq \int_{0}^{2}e^{-x^{2}}x^{3}\left(1+\frac{x^{2}}{2}+\frac{\frac{1}{6}x^{4}}{2}-\frac{1}{30\cdot24}x^{8}+\frac{1}{42\cdot720}x^{12}\right)dx+\int_{2}^{\infty}x^{5}\left(e^{-x^{2}}+e^{-4x^{2}}\right)dx$$

Proof :

We have as approximation :

$$\frac{1}{2}\int_{0}^{3}e^{-x}x\left(1+\frac{x}{2}+\frac{\frac{1}{6}x^{2}}{2}-\frac{1}{30\cdot24}x^{4}+\frac{1}{42\cdot720}x^{6}-\frac{1}{30\cdot8!}x^{8}\right)dx+\frac{1}{2}\int_{3}^{\infty}x^{2}\left(e^{-x}+e^{-2x}+e^{-3x}+e^{-4x}\right)dx<\zeta(3)$$

The proof can be done by hand .

The first integral is equal to : $$11/10+148087/(22400\exp\left(3\right))$$

The second integral is equal to :

$$1/2\left(85/(32e^{12})+101/(27e^{9})+25/(4e^{6})+17/e^{3}\right)$$

Now starting from my previous answer we have :

$$\left(3\int_{0}^{\left(2-\sqrt{3}\right)}\frac{x-\frac{x^{3}}{3}+\frac{2}{15}x^{5}+7x\left(-\frac{x^{3}}{3}+\frac{2}{15}x^{5}\right)^{2}}{x}dx-\frac{1}{4}\pi\ln\left(2-\sqrt{3}\right)-1\right)<C$$

The integral is equal to :

$$(151012882-87183909\sqrt{3})/7425$$

Now we need some approximation of roots,logarithm,$\pi$,$e$

To be continued.

Answer 5

Not a complete answer, but this approach relies on a less numerical (albeit there is still a lot of tedious algebra involved) method -- there may even be hope to compute a closed-form for $(2C-1)\zeta(3)$. The method of proof is similar in flavor to the famous $\frac{355}{113} > \pi$.

Let \begin{align*} I(p) &= \int_0^1 p(x)\frac{(-\log x)^{3/2}(2\frac{1-x}{1+x})^{1/2}}{x^2+1}dx \\ J(p) &= \int_0^1 p(x)\frac{(-\log x)(2\frac{1-x}{1+x})}{x^2+1}dx \\ K(p) &= \int_0^1 p(x)\frac{(-\log x)^2}{x^2+1}dx \end{align*} defined on $\mathbb{Q}^*=\{p \in \mathbb{Q}[x]: p(x) \ge 0\}$. By Cauchy-Schwarz, $I \le \sqrt{JK}$. One can then show, for any $k \in \mathbb{N}$, that \begin{align*} J(x^k) &= \alpha_1 + \alpha_2C + \alpha_3\pi^2 \\ K(x^k) &= \beta_1 + \beta_2\xi(3)+ \beta_3\pi^3 \end{align*} for rationals $\alpha_1, \alpha_2, \alpha_3, \beta_1, \beta_2, \beta_3$. My idea to prove the inequality in this post is then to identify a $p \in \mathbb{Q}^*$ such that \begin{align*} J(p) = k_1(2C-1) \qquad \text{and} \qquad K(p) = k_2 \zeta(3) \end{align*} for which then $(2C-1)\zeta(3) \ge (\frac{I(p)}{k_1k_2})^2 > 1$. This is guaranteed to work, because $\sqrt{JK} - I$ can be made arbitrarily close to 0 by considering $p(x) = x^n q(x)$ and letting $n\rightarrow \infty$.

[The reason why this works, and also the motivation behind choosing the integrands of the functions, is the equality condition of Cauchy-Schwarz $\|fg\|_2^2 \le \|f\|_2\|g\|_2$ is $f \propto g$; in this context, $-\log(x) \propto 2\frac{1-x}{1+x}$. The latter is obviously not true, but $\lim_{x\rightarrow 1} \frac{-\log x}{2\frac{1-x}{1+x}} = 1$, and the discrepant portions are made arbitrarily small by multipling by $x^n$.]

The trouble is now finding this $p$, which will involve computer algebra. I've attempted finding lower-ordered $p$'s and found \begin{align*} p(x) = x\left(\frac{3456}{7}x^4 + \frac{11736}{35}x^3 + \frac{34928}{105}x^2+ \frac{11736}{35}x - \frac{2336}{15}\right) \end{align*} which gives $I(p) = \frac{16}{3}(2C-1)$ and $J(p) = \zeta(3)$; unfortunately, $p \notin \mathbb{Q}^*$ since $p$ can be negative. But, I'm putting this method out there for others to find higher-degree polynomials (and in the original $\frac{355}{113} > \pi$ integral proof, the authors had to consider a degree-18 polynomial).

Answer 6

§1. No solution but just an idea suggested by the inverse of $\zeta(.)$ and not mentioned before here: take the Euler product.

Euler's famous formula reads in our case

$$\frac{1}{\zeta(3)}= \prod_{i>=1}(1-\frac{1}{p_i^3})$$

here $p_i$ is the $i$-th prime number ($p_1=2$, $p_2=3$, ...).

We can write

$$\frac{1}{\zeta(3)}= \prod_{i=1}^{17}(1-\frac{1}{p_i^3})\times\prod_{i>=18}(1-\frac{1}{p_i^3})<\prod_{i=1}^{17}(1-\frac{1}{p_i^3}) $$

so it is sufficient to show that

$$\prod_{i=1}^{17}(1-\frac{1}{p_i^3}) < 2C-1$$

I have chosen the sub product from numerical considerations.

Now this might help if we had a product form of $C$ ...

Sorry, my wisdom ends here.

§2. Numerics

Truncated series play a role in some of the solutions provided so far

Approximating $C$ and $\zeta$ in the product

$$p = (2C-1)\zeta(3)$$

by its naive series, i.e.

$$p(m) :=\left(\sum _{n=1}^{m} \frac{1}{n^3} \right)\left(2 \sum _{n=1}^{m} \frac{(-1)^{n+1}}{(2 n-1)^2}-1\right)$$

where $m$ is a truncation index, we find the point $m_1$ where $p(m_1)=1$ is $m_1=61$ for odd $m$ and $m_1=158$ for even $m$.