# Proving that: $9.9998\lt \frac{\pi^9}{e^8}\lt 10$?

I would like to use mathematical tools to prove that $$9.9998\lt \frac{\pi^9}{e^8}\lt 10$$

With an on-line calculator I got

$$\frac{\pi^9}{e^8}\approx 9.9998387978$$ But I do not know any good way to prove this. I failed to use Taylor expansion $$e^8 =\sum_{n=0}^{\infty}\frac{8^n}{n!}$$

Any idea?

• Well, it looks like taking natural logs helps. You want $\ln(9.9998)<9\ln (\pi)-8<\ln (10)$. Still numerical. – lulu Oct 25 '17 at 16:01
• It's looks nicer to proof $\,2980909<100 \cdot \pi^9<2980910\,$ and $\,298095<100 \cdot e^8<298096\,$ . :-) – user90369 Oct 25 '17 at 16:37
• That's funny. How did you discover such inequality? – Jack D'Aurizio Oct 25 '17 at 16:42
• @GuyFsone: nbubis' answer to the linked question has a structure that is similar to my answer here. The point is that the constant $e^{\pi}$ appears in the context of j-invariants and modular forms, hence it can be computed through fast-convergent numerical algorithms, recalling the AGM mean for the computation of complete elliptic integrals of the first kind. nbubis' answer is very satisfactory by my point of view, but I will take some time to think about an alternative approach. I am not sure to be able to do better. – Jack D'Aurizio Oct 25 '17 at 18:09
• Guy: Please stop making very trivial edits to get you post bumped to the front page. This example here, among many others: indicate very inappropriate behavior. – Namaste Nov 3 '17 at 18:30

My approach can be summarized as follows:

1. By a well-known fact about Euler numbers, $\pi^9$ is given by a rational multiple of the fast-converging series $\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^9}$;
2. By the generalized continued fraction for the hyperbolic cotangent function, it is pretty simple to produce accurate rational approximations of $e^8$;
3. The given inequality can be proved by producing accurate rational approximations for both $\pi^9$ and $e^8$ through the previous points, then by comparing them.

1. $$\pi^9 = \frac{203325460470370265464832}{6820919298826171875}\pm 7\cdot 10^{-5}$$
2. $$e^8 = \frac{6456755}{2166}\pm 4\cdot 10^{-8}$$
3. Done.
• Amazing! $(+1)^\infty$ – hypergeometric Oct 25 '17 at 17:09
• I wish everything was as simple as "1,2,3,done" for me (like it is for you). +1 – Paramanand Singh Oct 26 '17 at 4:01
• @hypergeometric: $+500>(+1)^\infty$. One can do it by start a bounty. – Jack Nov 5 '17 at 2:23
• Any idea here math.stackexchange.com/questions/2643216/… – Guy Fsone Feb 13 '18 at 11:51
• @Jack can one we give a bounty on a solved question? – Guy Fsone Feb 13 '18 at 14:16