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?

Solution 1:

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

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.

Solution 2:

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

Solution 3:

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 $$\sum_{n=1}^\infty\frac 1{F_n}=2.5+\sum_{n=4}^\infty\frac 1{F_n}\\ \lt 2.5+\frac 1{0.94}\sum_{n=4}^\infty\frac {\sqrt 5}{\phi^n}\\ =2.5+\frac {\sqrt 5}{0.94\phi^3(\phi-1)}\lt 2.5+0.9087=3.4087\lt 4$$

Solution 4:

Another attempt: separate the series into two partial series: $$ \small \begin{array} {r|r} 1 & 1 \\ 1/2 & 1/3 \\ 1/5 & 1/8 \\ 1/13 & 1/21 \\ 1/34 & 1/55 \\ ... & ... \\ s_1 & s_2 \end{array} $$ Each sum $s_1,s_2$ is obviously smaller than $1,1/2,1/4,1/8,...$ (easily provable considering two steps in the Fibonacci-sequence) so the sum must be smaller than $2 \times (1+1/2+1/4+...) = 4 $

Solution 5:

Inverse Fibonacci sequence:

$$\frac{1}{1}, \frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \frac{1}{5}\ldots$$

Geometric series beginning with $1$ and ratio of $3/4$:

$$\frac{1}{1}, \frac{3}{4}, \frac{9}{16}, \frac{27}{64}, \frac{81}{256}, \ldots$$

Summing, the latter yields a series that is greater than the former; moreover, observe that

$$\sum_{n \geq 0} \Big(\frac{3}{4}\Big)^n = 4$$

thereby establishing the desired inequality.

Details that remain to fill in: Show that everything converges. Prove that the effect of the first couple terms will not be disastrous.