Setting the finite sum of squares equal to a square number yields only one non-trivial solution, when $n= 24$; the sum becomes $4900$, which is $70^2$ $$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$$ https://youtu.be/vzjbRhYjELo?t=4m21s
The infinite sum of reciprocal squares is equivalent to evaluating the Riemann Zeta Function at $s=2$. $Zeta(2) = \frac{\pi^2} 6$
https://www.youtube.com/watch?v=d-o3eB9sfls
If you replace $\pi$ with $\tau (\tau = 2*\pi = 6.2931...)$ we can see that $Zeta(2)$ equals $\frac {(\tau^2)}{24}$
$24$ appears in both summations of squares. One way is a finite sum starting at the apex of a square pyramid toward an infinitely far away base, and the other problem involves starting at the base and going up towards an infinitely far away apex.
What is happening with the number $24$ and this "inversion" of the summations of these square pyramids?!