Edit: I saw that Math.SE question and talked/linked it in my last paragraph as well as in the comments.
Here are two series for $\pi^2$
$$\frac{\pi^2}{6} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{25} + \frac{1}{36} ... $$
$$\frac{\pi^2}{8} = 1 + \frac{1}{9} + \frac{1}{25} + \frac{1}{49} + \frac{1}{81}...$$
These are the simplest and easiest series for pi that I know. They're all positive terms, they're all simple fractions of squares.
What I'd like to know is, can either of those two series be proved purely geometrically? Preferably involving only squares and circles.
I have seen a video proving the first one using polynomials and trig series. I also saw another video proving a related (but still simple) series using an integral and the arctan function. None of them were very geometric.
It seems tantalizing because the series only involves fractions of squares (numerical squaring), so you'd hope there was some proof involving squares (the shape). For reference, $c^2 = a^2 + b^2$ has a purely geometrical proof by rearranging triangles into 2 squares (a square with a square hole in the middle), and then just a little algebra.
Btw, I'm aware that "squaring the circle" is impossible. I'm not trying to do that. I'm not trying to construct (with compass and straightedge) a square whose area is equal to a circle. I could easily draw a square, label the sides with $\sqrt{\pi}$ or $\pi$, and then prove the area is $\pi$ or $\pi^2$, not using compass nor straightedge. I could just as easily draw a circle, label its circumference as 1, then draw a bounding square and prove its area is $\frac{1}{\pi^2}$.
What I'm actually trying to do is show that a unit square plus one quarter of it plus one ninth of it, etc, is somehow equal to a one sixth of a square whose side length is $\pi$. I suppose this could be done by involving a circle of unit diameter, or a circle of unit circumference whose diameter is therefore 1/pi.
Alternatively, I'm trying to show that a unit square plus one ninth of it plus one twenty-fifth of it, etc, is somehow equal to one eighth of a square whose side length is pi.
Using triangles is okay too but preferably without having to use the trig functions or their infinite series. Hopefully there will be a way to use area of a triangle (bh/2) to get to squares.
I've also looked at Basel's Problem and the last proof which is claimed to be "the most elementary", but it involves binomial theorem and vieta's formula and furthermore does not seem to involve any squares (the shape).
I've also seen this question on Math.SE, but the first answer has a dead link and the pictures seem to involve a stack of squares and some curved line related to e, not pi and not a circle.
