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 $