<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] First, note that $\frac{1+\sqrt{5}}{2}$ is a root of the quadratic equation
$$f(x)=x^2-x-1$$
This equation is increasing on $(1/2,\infty)$ which is easily shown as
$$f'(x)=2x-1$$
which has one zero at $x=\frac{1}{2}$ and $f'(1)=1>0$. Now, note that
$$f\left(\frac{81}{50}\right)=\left(\frac{81}{50}\right)^2-\left(\frac{81}{50}\right)-1=\frac{11}{2500}>0.$$
Since $81/50>1/2$, and $f(81/50)>0$, we can conclude that 
$$\frac{81}{50}>\frac{1+\sqrt{5}}{2}.$$
However, if we take the first $40$ terms of the infinite sum, we get
$$\frac{\pi^2}{6}=\sum_{n=1}^\infty \frac{1}{n^2}>\sum_{n=1}^{40} \frac{1}{n^2}=\frac{46252969210499754415427421586309}{28546916554875489385168794240000}=1.62024>1.62=\frac{81}{50}>\frac{1+\sqrt{5}}{2}.$$
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