# Proving that $7<\frac{5\phi e}{\pi}< 7.0000689$ where $\phi$ is the Golden Ratio

In my numerics I need to show the following approximation:

$$7<\frac{5\phi e}{\pi}< 7.000069,$$

where $\phi = \dfrac{1+\sqrt{5}}{2}$ is the Golden Ratio and $e$ is the Euler's constant as well.

Using an online calculator I got $$\color{red}{\frac{5\phi e}{\pi}\approx 7.000068914.}$$

I have tried using the Taylor series

$$e=\sum_{n=0}^{\infty}\frac{1}{n!},$$

But This is not enough to approximate accurately the whole term.

Does anyone have an idea to approximate $\frac{5\phi e}{\pi}$ ? Some other methods have been used here How to prove that: $19.999<e^\pi-\pi<20$? and here Proving that: $9.9998\lt \frac{\pi^9}{e^8}\lt 10$?

Mimicking the approach shown in the mentioned question, both the continued fractions of $\phi$ and $e$ are simple and well-known. About $\frac{1}{\pi}$, one may use Ramanujan's fast convergent series to get $$\frac{1}{\pi}\approx \frac{13562078735077}{30127364780832 \sqrt{2}}$$ with a relative error which is less than $3\cdot 10^{-8}$. Since $e\approx\frac{23225}{8544}$ with a relative error which is less than $3\cdot 10^{-9}$ and $\phi\approx \frac{4181}{2584}$ with a relative error which is less than $5\cdot 10^{-8}$, we have $$\frac{5\phi e}{\pi}\approx7+\frac{1}{14690}$$ with a relative error which is less than $10^{-7}$, proving the claim.

$\phi$ is simply $(1 + \sqrt 5)/2$. Square roots are easily approximable using Newton's method.

For $\pi$ you have lots and lots of very fast algorithms.

Part 1: $\frac{1}{\pi}$

Archimedes obtains: $\frac{223}{71} < \pi<\frac{22}{7}$
Hence: $\frac{7}{22} <\frac{1}{\pi}<\frac{71}{223}$
We'll also use a trick: $\pi\approx 3\to\frac{1}{\pi}\approx\frac{7}{21}$

Part 2: $\phi$

$\phi=\frac{1+\sqrt{5}}{2}\approx 1.618$

The golden ratio is connected to the Fibonacci numbers: 1,1,2,3,5,8,13,21,34,55,89…

And notably: $\phi=\displaystyle\lim_{n \to \infty} \frac{F_{n+1}}{F_n}$

Here they are the first approximations:$\frac{2}{1};\frac{3}{2};\frac{5}{3};\frac{8}{5};\frac{13}{8};\frac{21}{13};\frac{34}{21};\frac{55}{34};\frac{89}{55}...$

We'll consider the intervals of: $\frac{21}{13}<\phi<\frac{34}{21}$ and $\frac{55}{34}<\phi<\frac{89}{55}$

Part 3: Euler’s constant

As: $e=\displaystyle\sum_{n=0}^{\infty}\frac{1}{n!}=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...$

$e=2.5+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}+\frac{1}{720}+...$

OK, $e$ is bigger than 2.6 (because 1/6 is evident bigger than 1/10) :

$2.6=\frac{26}{10}=\frac{13}{5}$

Euler’s number is also smaller than 2.72 (no proof ; and 2.75 is „too” big):

$2.72=2\frac{72}{100}=2\frac{18}{25}=\frac{68}{25}$

Finally

$5\phi\frac{1}{\pi}e$ :

-Lower part:
Version 1 (with the trick): $5*\frac{21}{13}*\frac{7}{21}*\frac{13}{5}=7$

Version 2: $5*\frac{55}{34}*\frac{7}{22}*\frac{68}{25}=5*\frac{5*11}{34}*\frac{7}{2*11}*\frac{2*34}{5*5}=7$

-Biggest part: Our convergence will be slow – we reduce into three tries:

First try: $5*\frac{34}{21}*\frac{7}{22}*\frac{68}{25}=5*\frac{34}{3*7}*\frac{7}{2*11}*\frac{2*34}{5*5}=\frac{34^2}{3*5*11}=\frac{1156}{165}=7.00(60)\approx 7.0061$

Second try: $5*\frac{55}{34}*\frac{71}{223}*\frac{68}{25}=\frac{2*11*71}{223}\approx 7.0045$

Third try: $5*\frac{89}{55}*\frac{7}{22}*\frac{68}{25}=\frac{7*34*89}{(5*11)^2}\approx 7.0023$