<s>[INST] How can I prove analytically that the golden ratio is less then $\frac{\pi^2}{6}$. Or stated in other terms, prove that
$$\frac{1+\sqrt{5}}{2} < \sum_{n=1}^{\infty}\frac{1}{n^2}$$
 [/INST] $\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}=\frac12\sum_{n=1}^\infty \frac1{n^2}$ (see here). Now, the inequality is equivalent to $\frac{1+\sqrt5}4<\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}.$
\begin{equation}
\begin{split}
\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}&=\sum_{k=1}^\infty\left(\frac1{(2k-1)^2}-\frac1{(2k)^2}\right)\\
&>\left(\frac1{1^2}-\frac1{2^2}\right)+\left(\frac1{3^2}-\frac1{4^2}\right)+\left(\frac1{5^2}-\frac1{6^2}\right)\\
&>\frac{1+\sqrt5}4.\\
\end{split}
\end{equation}
</s>