How can I prove analytically that the golden ratio is less then $\frac{\pi^2}{6}$. Or stated in other terms, prove that
$$\frac{1+\sqrt{5}}{2} < \sum_{n=1}^{\infty}\frac{1}{n^2}$$

Without needing to know the numerical value of $\pi^2/6$, you can use the infinite series representation of the golden ration and some very crude bounds to arrive at the conclusion. Recall $$\varphi = \sum^\infty_{n=0} \frac{(-1)^n}{F_n F_{n+1}}$$ where $F_0 = 1, F_1 = 1$ and $F_n = F_{n-1} + F_{n-2}, \,\,\, n \ge 2$ is the Fibonacci sequence. Now you can prove the inequality term-by-term but using each pair of consecutive terms in the sum: $$\frac{1}{F_nF_{n+1}} - \frac{1}{F_{n+1}F_{n+2}} < \frac{1}{(n+1)^2} + \frac{1}{(n+2)^2}, \,\,\,\, n = 0,2,4,\cdots. \,\,\,\,\,\,\,\,\,\,\, (1)$$ Indeed, $$\frac{1}{F_nF_{n+1}} - \frac{1}{F_{n+1}F_{n+2}} =  \frac{F_{n+2} - F_n}{F_nF_{n+1}F_{n+2}} = \frac{1}{F_n{F_{n+2}}},$$ and we have $F_n \ge n$, so $$\frac{1}{F_n F_{n+2}} \le \frac{1}{n(n+2)} \le \frac{1}{2n^2} + \frac{1}{2(n+2)^2} < \frac {1}{(n+1)^2} + \frac{1}{(n+2)^2}.$$ Note, the last inequality only necessarily holds when $2n^2 \ge (n+1)^2$ (i.e, $n \ge 3$). You can verify $(1)$ when $n=0,2$ separately.
