What's the best coefficient to have $b^{-x^{a}}+c-\varepsilon<\sum_{n=1}^{\infty}\left(-1\right)^{n}m\left(x^{n}\right)

Question

Hi it's the last follow up of Show that $\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}-\cos\left(\frac{2}{x}\right)>0$ for $x\geq 4/\pi$

I wrote :

Bonus/conjecture :

Let $x\in[\pi/3,3/2]$ then define :

$$m\left(x\right)=\cos\left(\frac{2}{x}\right)-\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}$$

Then it seems $\exists a,b,c\in[-\infty,\infty]$ and $\exists \varepsilon ,0<\varepsilon$ such that :

$$b^{-x^{a}}+c-\varepsilon<\sum_{n=1}^{\infty}\left(-1\right)^{n}m\left(x^{n}\right)<b^{-x^{a}}+c$$

It's slightly modified to give much result , currently we have nice result with :

$$a=\frac{95}{100},b=\frac{3886}{1000},c=\frac{-49}{1000}$$

Or : $$b=e^{\frac{e}{2}},a=\frac{\exp\left(2-\frac{e}{2}\right)}{2},c=-\exp\left(-\exp\left(\frac{e^{2}}{2-e+e^{2}}\right)\right)$$

Stopping the infinite series to $200$ summations .

Question :

What's the "best" (which cannot be replaced by a smaller or greater one) coefficient to have the inequality ?

Answer 1

I let a possible series which converges :

Let :

$$m\left(x\right)=\cos\left(\frac{2}{x}\right)-\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}},f\left(x\right)=\sum_{n=1}^{100}\left(-1\right)^{n}m\left(x^{n}\right),b=e^{\frac{e}{2}},a=\frac{\exp\left(2-\frac{e}{2}\right)}{2},c=-\exp\left(-\exp\left(\frac{e^{2}}{2-e+e^{2}}\right)\right)$$

Then define :

$$h\left(x\right)=\left|f\left(x\right)-\left(b^{-x^{a}}+c\right)\right|$$

$$j\left(x\right)=\left|h\left(x\right)-\left(\frac{\left(h\left(\frac{\pi}{3}\right)-h\left(1+e^{-1}\right)\right)}{\frac{\pi}{3}-1-e^{-1}}\left(x^{2}-\left(1+e^{-1}\right)^{2}\right)+h\left(1+e^{-1}\right)\right)\right|$$

$$t\left(x\right)=\left|j\left(x\right)-\left(\left|\frac{\left(j\left(\left(1+e^{-1}\right)\right)-j\left(\sqrt{\frac{\pi}{3}\left(1+e^{-1}\right)}\right)\right)}{\left(1+e^{-1}\right)-\sqrt{\frac{\pi}{3}\left(1+e^{-1}\right)}}\left(x^{2}-\frac{\pi}{3}\left(1+e^{-1}\right)\right)+j\left(\sqrt{\frac{\pi}{3}\left(1+e^{-1}\right)}\right)\right|\right)\right|$$

$$r\left(x\right)=\left|t\left(x\right)-\left(\left|\frac{\left(t\left(\left(1+e^{-1}\right)\right)-t\left(\sqrt{\frac{\pi}{3}\left(1+e^{-1}\right)}\right)\right)}{\left(1+e^{-1}\right)-\sqrt{\frac{\pi}{3}\left(1+e^{-1}\right)}}\left(x-\sqrt{\frac{\pi}{3}\left(1+e^{-1}\right)}\right)+t\left(\sqrt{\frac{\pi}{3}\left(1+e^{-1}\right)}\right)\right|\right)\right|$$

$$k\left(x\right)=\left|r\left(x\right)-\left(\left|\frac{\left(r\left(\left(1+e^{-1}\right)\right)-r\left(\sqrt{\left(1+e^{-1}\right)\sqrt{\frac{\pi}{3}\left(1+e^{-1}\right)}}\right)\right)}{\left(1+e^{-1}\right)-\sqrt{\left(1+e^{-1}\right)\sqrt{\frac{\pi}{3}\left(1+e^{-1}\right)}}}\left(x-\left(1+e^{-1}\right)\right)+r\left(1+e^{-1}\right)\right|\right)\right|$$

And so on ...