First solution:
![\begin{equation} \label{eq:poly} \sum_{n=1}^\infty \frac{1}{n^2} = \int_1^\infty \frac{1}{x^2}\>\> dx \end{equation} $$\int_1^\infty \frac{1}{x^2}\>\>\>dx=\lim_{k\to \infty}\int_1^k \frac{1}{x^2}\> dx$$ \begin{equation} =\lim_{k\to \infty}[\left[ -\frac{1}{x} \right]_1^k] \end{equation} \begin{equation} = \lim_{k\to \infty}[-\frac{1}{k}]+1=0+1=1 \end{equation} Therefore, $$ \sum_{n=1}^\infty \frac{1}{n^2} = 1 $$ Second solution: $$\sum_{n=1}^\infty \frac{1}{n^2} = \left \frac{\pi^2}{6} \right $$](https://mako-sawin.com/wp-content/ql-cache/quicklatex.com-5b968809019891eeff1a6ebc7d0063c2_l3.png "Rendered by QuickLaTeX.com")
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