Problem :
Show that :
$$\frac{1}{\zeta(3)}<2C-1$$
Where we can see the zeta function and the Catalan's constant .
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
Best Answer
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 $$\log(\Delta_{p+1}-\Delta_p)\sim 4.45 -5.22 \, p$$