### Testing

Posted:

**Mon Sep 28, 2020 8:49 pm**\displaystyle\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}:

\(\displaystyle\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}\)

\underset{n=1}{\overset{\infty}{\Sigma}} \frac{1}{n^2}=\frac{\pi^2}{6}:

\(\underset{n=1}{\overset{\infty}{\Sigma}} \frac{1}{n^2}=\frac{\pi^2}{6}\)

\sqrt{6 \underset{n=1}{\overset{\infty}{\Sigma}} \frac{1}{n^2}}=\pi:

\(\sqrt{6 \underset{n=1}{\overset{\infty}{\Sigma}} \frac{1}{n^2}}=\pi\)

The square root is not well rendered.

\(\displaystyle\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}\)

\underset{n=1}{\overset{\infty}{\Sigma}} \frac{1}{n^2}=\frac{\pi^2}{6}:

\(\underset{n=1}{\overset{\infty}{\Sigma}} \frac{1}{n^2}=\frac{\pi^2}{6}\)

\sqrt{6 \underset{n=1}{\overset{\infty}{\Sigma}} \frac{1}{n^2}}=\pi:

\(\sqrt{6 \underset{n=1}{\overset{\infty}{\Sigma}} \frac{1}{n^2}}=\pi\)

The square root is not well rendered.