Prove that ${e\over {\pi}}\lt{\sqrt3\over{2}}$ without using a calculator.

Question

I have been working on a known question for a long time (this is "Proving that $e^{\pi}-{\pi}^e\lt 1$ without using a calculator") during this time I realized the ${e\over {\pi}}\lt{\sqrt3\over{2}}$.
I have no solution so far. Do you have any idea?

Answer 1

Calculation without any computer, only with patience. :-D

$\displaystyle e<\frac{\sqrt{3}}{2}\pi\enspace$ is equivalent to $\enspace\displaystyle \frac{2}{9}\sum\limits_{n=0}^\infty \frac{2^n}{n!} =\frac{2e^2}{9}<\frac{\pi^2}{6}=\zeta(2)$

Case $\,(A)\,$ :

We have $\enspace\displaystyle \prod\limits_{k=0}^n \frac{k+9}{4}>1\enspace$ for all $\,n\geq 0\,$ and therefore $\,4^{n+1}8!<(n+9)!\,$ .

It follows $\enspace\displaystyle \frac{2^{n+1}8!}{(n+9)!}<\frac{1}{2^{n+1}}\enspace$ and with the summation over $\,n=0,1,2,…\,$ we get

$\displaystyle \sum\limits_{n=0}^\infty\frac{2^{n+1}8!}{(n+9)!}< \sum\limits_{n=0}^\infty \frac{1}{2^{n+1}}=1\enspace$ which leads to $\enspace\displaystyle e^2 = \sum\limits_{n=0}^\infty \frac{2^n}{n!} < \sum\limits_{n=0}^7 \frac{2^n}{n!} + \frac{2^9}{8!} \,$ .

Case $\,(B)\,$ :

The Euler–Maclaurin formula (https://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula) gives us a good approximation for $\,\zeta(2)\,$ and we get

$\displaystyle \sum\limits_{n=0}^4 \frac{1}{n^2} + \frac{1}{5} + \frac{1}{50} + \frac{1}{6\cdot 125} - \frac{1}{30\cdot 5^5}<\zeta(2)\,$ .

Cases $\,(A)\,$ and $\,(B)\,$ together :

Now we have to prove the middle part of the following inequality.

$\displaystyle \frac{2e^2}{9} = \frac{2}{9}\sum\limits_{n=0}^\infty \frac{2^n}{n!} < \frac{2}{9}\left(\sum\limits_{n=0}^7 \frac{2^n}{n!} + \frac{2^9}{8!}\right)<$

$\displaystyle <\sum\limits_{n=0}^4 \frac{1}{n^2} + \frac{1}{5} + \frac{1}{50} + \frac{1}{6\cdot 125} - \frac{1}{30\cdot 5^5}<\zeta(2)=\frac{\pi^2}{6}\,$

or simply $\enspace\displaystyle \frac{2e^2}{9} < \frac{4658}{2835} < \frac{3701101}{2250000} < \frac{\pi^2}{6}$

And with $\,4658\cdot 2250000 =10480500000 < 10492621335=2835\cdot 3701101$

the claim is proofed.