How to prove that: $19.999<e^\pi-\pi<20$?

Question

I would like to know how to prove $$e^\pi-\pi\sim 20.$$ More precisely, I want to show by using only mathematical tools that, $$19.999<e^\pi-\pi<20$$

---

I have checked with online calculator and I got $$e^\pi-\pi\approx19.9990999792\sim 20.$$

I tried to use the Tyalor expansion for exponential $$ e^\pi =\sum_{n=0}^{\infty} \frac{\pi^n}{n!} = \pi +1+\sum_{n=2}^{\infty} \frac{\pi^n}{n!}$$ then, $$e^\pi -\pi =1+\sum_{n=2}^{\infty} \frac{\pi^n}{n!}$$ which is not easy to continue from here, since the factor $\pi^n$ is involved. Any idea?

Answer 1

Use the iteration to calculate Gelfond's constant: $$k_0 = 1/\sqrt{2},\quad k_{n+1}={\frac {1-{\sqrt {1-k_{n}^{2}}}} {1+{\sqrt {1-k_{n}^{2}}}}}$$ $$e^\pi =\lim_{n\to\infty} \left(\frac{4}{k_{n+1}}\right)^{2^{-n}}$$

Within two iterations you should have (assuming you know $\pi$ to 5 or more digits and that you're willing to compute three square roots), that $e^\pi-\pi\approx 19.999$:

$$k_1=3-2\sqrt{2}$$ $$k_2=33+24\sqrt{2}-4\sqrt{140+99\sqrt{2}}$$ And: $$\sqrt{4\over k_2} \approx 19.99926 + \pi$$