Proving ${\sum_{n=1}^\infty {1\over F_n}} <4$ I'm trying to prove the sum of Fibonacci numbers' reciprocals is less than 4, which is:
$${\sum_{n=1}^\infty {1\over F_n}} <4$$
It makes me confused because the only information I know about Fibonacci numbers that might be useful are its recurrence relation and general formula. But when dealing with reciprocals, I found the info hard to use.
I also thought of induction: maybe turning this into:
 $${\sum_{n=1}^\infty {1\over F_n}} <4-A$$
where A is related to $F_n$. But this method also seems to be not working.
Could anyone please give me some hints?

This solution uses only simple arithmetic and some recursion.
First we prove that $F_{n+1}-\dfrac{3}{2}F_n\gt 0$ for all $n\ge 4$. Simplify the left hand part:
$$
F_{n+1}-\dfrac{3}{2}F_n = F_{n}+F_{n-1}-\dfrac{3}{2}F_n = F_{n-1}-\dfrac{1}{2}F_n = F_{n-1}-\dfrac{1}{2}(F_{n-1}+F_{n-2}) = \dfrac{1}{2}(F_{n-1}-F_{n-2}) = \dfrac{1}{2}F_{n-3}
$$
Since $F_{n-3}$ is positive for all $n\ge 4$, the statement is true.
Now rearrange the inequality: $F_{n+1}\gt\dfrac{3}{2}F_n$
Invert both sides: $\dfrac{1}{F_{n+1}}\lt\dfrac{2}{3}\dfrac{1}{F_n}$
This holds for all $n\ge 4$, so we can recursively expand the right-hand side:
$$
\dfrac{1}{F_{n+1}}\lt\dfrac{2}{3}\dfrac{1}{F_n}\lt\left(\dfrac{2}{3}\right)^2 \dfrac{1}{F_{n-1}}\lt\cdots\lt\left(\dfrac{2}{3}\right)^{n-3} \dfrac{1}{F_4}
$$
This provides an upper bound for the sum since each term is smaller than a corresponding term in a convergent geometric series. It could also be used as an alternative proof of convergence for Barry Cipra's excellent answer, but using finite descent rather than induction.
The rest just involves calculating this upper bound. First a partial sum of $\dfrac{1}{F_n}$ where $n>4$:
$$
\sum\limits_{n=5}^\infty \dfrac{1}{F_n} \lt \sum\limits_{k=1}^\infty \left(\dfrac{2}{3}\right)^{k} \dfrac{1}{F_4}=\dfrac{1}{3}\cdot\dfrac{2}{3}\sum\limits_{k=0}^\infty \left(\dfrac{2}{3}\right)^{k} = \dfrac{2}{9}\cdot\dfrac{1}{1-\frac{2}{3}} = \dfrac{2}{3}
$$
Then to complete the proof we add the first four terms to both sides:
$$
\sum\limits_{n=1}^\infty \dfrac{1}{F_n} = \dfrac{1}{F_1}+\dfrac{1}{F_2}+\dfrac{1}{F_3}+\dfrac{1}{F_4}+\sum\limits_{n=5}^\infty \dfrac{1}{F_n} \lt 1+1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{3} = \frac{7}{2} \lt 4
$$
