1

Challenging problem :

Show that :

$$\frac{\gamma}{2-\gamma}\pi-\ln(3+\gamma)<0$$

I come up with this inequality with What is the limit $\lim_{x\to 0}\left(\frac{x!x!!!x!!!!!...}{x!!x!!!!x!!!!!!...}\right)^{-\frac{1}{x}}=^?$

I can show the inequalities :

$$\frac{\pi}{2}-1<\gamma<\frac{1}{\sqrt{3}}$$

Using the Leibniz formula (expansion of arctangent) and the first harmonic number using a computer . It doesn't help here

We can also use continued fraction for $\pi$ and for $\gamma$ see https://mathworld.wolfram.com/Euler-MascheroniConstantContinuedFraction.html

We can also continue to get :

$$\frac{\gamma}{2-\gamma}\pi-\ln\left(3+\gamma\right)+\sqrt{2}\ln\left(1+\left(1+\gamma-\frac{\pi}{2}\right)^{2}\right)-\frac{1}{2}\ln\left(1+\left(1+\gamma-\frac{\pi}{2}\right)^{3}\right)>0$$

How to show it by hand without a computer ?

Barackouda
  • 3,879
  • We have also $$\frac{\gamma}{2-\gamma}\pi-\ln\left(3+\gamma\right)+\sqrt{2}\ln\left(1+\left(1+\gamma-\frac{\pi}{2}\right)^{2}\right)-\frac{1}{2}\ln\left(1+\left(1+\gamma-\frac{\pi}{2}\right)^{3}\right)-\frac{1}{3}\ln\left(1+\left(1+\gamma-\frac{\pi}{2}\right)^{4}\right)-6\ln\left(1+\left(1+\gamma-\frac{\pi}{2}\right)^{5}\right)-70\ln\left(1+\left(1+\gamma-\frac{\pi}{2}\right)^{6}\right)\simeq 0$$ – Barackouda Sep 09 '22 at 13:47
  • This is a hard one. It is fairly easy to establish $$3<\pi<\frac{22}{7}$$ And $$\frac{39}{72}<\gamma<\frac{49}{72}$$ Without using any software, but I don't think these bounds are tight enough. – K.defaoite Sep 09 '22 at 16:30
  • @K.defaoite Thanks for the reply . What about continued fraction ? – Barackouda Sep 09 '22 at 16:34
  • Be more specific. – K.defaoite Sep 09 '22 at 16:36
  • @K.defaoite See the bountied answer here https://math.stackexchange.com/questions/4330847/showing-sqrt-frace2-cdot-frace-pi-left-frace2-frac1e-right – Barackouda Sep 09 '22 at 16:48

0 Answers0