Basel Problem - Area of $\frac 16$ of Circle with Radius $\sqrt{\pi}$. There are several proofs to the solution of the well-known Basel Problem, i.e.
$$\sum_{n=1}^\infty \frac 1{n^2}=\frac {\pi^2}6$$

Is is possible to create a geometrical interpretation of this identity in the form of the area of $\frac 16$ of a circle with radius $\sqrt{\pi}$?



Topic: Radius $\sqrt{\pi}$

Today (and this gonna be my last post- unless I get requests) with the golden ratio:

$\phi=\frac{1+\sqrt{5}}{2}$ ; Property : $\phi^2=\phi+1$

I found this pythagorean relation:
$$
(\sqrt{\pi})^2\leq\phi^2+(\frac{\phi}{4})^2+0.6^2
$$

Accuracy:
$$$$ \begin{align} \frac{17}{16}\phi^2+0.36\approx3.141661\\    \pi\approx3.141593 \\  { For-comparison:}
\frac{22}{7}\approx3.14\color{red}{2857} \\  \end{align}
