How can I prove analytically that the golden ratio is less then $\frac{\pi^2}{6}$.

Question

Or stated in other terms, prove that $$\frac{1+\sqrt{5}}{2} < \sum_{n=1}^{\infty}\frac{1}{n^2}$$

Answer 1

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.

Answer 2

\begin{align} \frac{1+\sqrt{5}}{2}\lt\frac{\pi^2}{6} \iff&3+3\sqrt{5}\lt\pi^2\\ \iff&3\sqrt{5}\lt\pi^2-3\\ \iff&(3\sqrt{5})^2\lt(\pi^2-3)^2\quad(\text{both sides are positive})\\ \iff&45\lt\pi^4-6\pi^2+9\\ \iff&\pi^4-6\pi^2-36\gt0\\ \end{align} Using, for example, $3.14\lt\pi\lt3.15$ we have that $$\pi^4-6\pi^2-36\gt3.14^4-6\cdot3.15^2-36\gt0$$ which implies that $$\frac{1+\sqrt{5}}{2}\lt\frac{\pi^2}{6}$$

Answer 3

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

$$\frac{\pi^2}{6}=\sum_{n=1}^\infty \frac{1}{n^2}>\sum_{n=1}^{40} \frac{1}{n^2}=\frac{46252969210499754415427421586309}{28546916554875489385168794240000}=1.62024>1.62=\frac{81}{50}>\frac{1+\sqrt{5}}{2}.$$

Answer 4

$\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}=\frac12\sum_{n=1}^\infty \frac1{n^2}$ (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}