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?
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Peter
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JV.Stalker
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1Possible duplicate of A question comparing $\pi^e$ to $e^\pi$ – caverac Sep 22 '17 at 12:00
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1With "prove" you mean a method not involving calculating $e$, $\pi$ or other constants by hand, right ? – Peter Sep 22 '17 at 12:04
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3Possible duplicate of Comparing $\pi^e$ and $e^\pi$ without calculating them – gen-ℤ ready to perish Sep 22 '17 at 12:06
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Please read the question, it's not a duplicate – orangeskid Sep 22 '17 at 12:32
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Peter, yes. Same as you can see on the following link . https://math.stackexchange.com/questions/1410230/proving-that-e-pi-pie-lt-1-without-using-a-calculator – JV.Stalker Sep 22 '17 at 12:33
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this will be tricky, as it's a close inequality. $pi/e \approx 0.865256$ while $\sqrt{3}/2 \approx 0.866205$. – Michael Lugo Sep 22 '17 at 13:38
1 Answers
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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.
user90369
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I am sorry for the late congratulation. Your solution should be taught. Thanks for it. – JV.Stalker Sep 23 '17 at 08:45
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