Infinite Circles 2. Now with roots

Level 1

$\sqrt{1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}+\cdot\cdot\cdot} \times\sqrt{6}$

Find the closed form of the expression to at least 3 decimal places.

1 solution

Tom Engelsman
May 5, 2021

We have:

$\Large \sqrt{6 \cdot \Sigma_{n=1}^{\infty} \frac{1}{n^2}} = \sqrt{6 \cdot \zeta(2)} = \sqrt{6 \cdot \frac{\pi^2}{6}} = |\pi| = \boxed{\pi}.$