Proving ${\sum_{n=1}^\infty {1\over F_n}} <4$

$$S={1\over1}+{1\over1}+{1\over2}+{1\over3}+{1\over5}+{1\over8}+\cdots=2+{1\over1+1}+{1\over1+2}+{1\over2+3}+{1\over3+5}+\cdots\\\lt2+{1\over1+1}+{1\over1+1}+{1\over2+2}+{1\over3+3}+\cdots\\ =2+{1\over2}\left({1\over1}+{1\over1}+{1\over2}+{1\over3}+\cdots \right)=2+{1\over2}S$$

so ${1\over2}S\lt2$, or $S\lt4$.

Remark: As B. Mehta points out, this argument only works if the sum converges. So here's a cheap way to show convergence. By induction, if $F_n\ge cn^2$ and $F_{n-1}\ge c(n-1)^2$, which is true for $n\lt4$ if $c$ is sufficiently small, then

$$F_{n+1}=F_n+F_{n-1}\ge c(n^2+(n-1)^2)=c(n^2+(n^2-2n)+1)\ge c(n^2+2n+1)=c(n+1)^2$$

since $n^2-2n\ge2n$ if $n\ge4$. It follows that $\sum{1\over F_n}\le{1\over c}\sum{1\over n^2}$, which converges.

Prove by induction that $$2^n\leq F_{2n}$$ and $$2^{n}\leq F_{2n+1}$$

We know the Fibonacci series is very close to geometric, so we can sum the reciprocals of a similar series as an upper bound. Recall Binet's formula $$F_n=\frac {\phi^n-\psi^n}{\sqrt 5}$$ where $\phi=\frac 12(1+\sqrt 5)\approx 1.618, \psi=\frac 12(1-\sqrt 5)\approx -0.618$

The first three terms of the inverse Fibonacci series are $\frac 11+\frac 11 + \frac 12=2.5$. After that we have $|\psi^n| \lt 0.03 \phi^n$, so $\frac 1{F_n} \lt \frac 1{0.94\phi^n}$ so $$\sum_{n=1}^\infty\frac 1{F_n}=2.5+\sum_{n=4}^\infty\frac 1{F_n}\\ \lt 2.5+\frac 1{0.94}\sum_{n=4}^\infty\frac {\sqrt 5}{\phi^n}\\ =2.5+\frac {\sqrt 5}{0.94\phi^3(\phi-1)}\lt 2.5+0.9087=3.4087\lt 4$$