[Solution] Sum of reciprocal squares.

Problem was


1) Show that $\frac{\pi}{4} \gt \sqrt{2-\sqrt{2}}$

2) Show that $\sum_{n=2}^{\infty} \frac{1}{n^2} \gt \frac{3}{5}$


For part 1) we can give a geometric proof of sorts.

A circle of radius 1. The length of arc AB is $\frac{\pi}{4}$. The length of the line segment AB is $2 \sin \frac{\pi}{8} = \sqrt{2-\sqrt{2}}$. Since arc length $\gt$ line segment length, we are done.




For part 2)


We use the following inequality

$$ n^2 \lt \left(n - \frac{1}{3}\right) \left(n + \frac{2}{3}\right)$$


This gives us

$$ \frac{1}{n^2} \gt \frac{1}{n - \frac{1}{3}} - \frac{1}{n + \frac{2}{3}}$$

Setting $n = 2, 3, \dots$ and adding we get a telescoping series on the right and the only term that remains is $\frac{1}{2 - \frac{1}{3}} = \frac{3}{5}$.

This implies part $1$ because of the classic result that $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$.