Sum of reciprocals of Fibonacci numbers convergence I am trying to prove the convergence/divergence of the series´
 $$\sum_{i=1}^\infty \frac{1}{F_n} = 1+1+\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{8}+...$$
${F_n}$ being the Fibonacci sequence.
The Fibonacci sequence is defined without recursion by:
$${F_n}=\frac{\phi^n-(-\phi)^{-n}}{\sqrt{5}} \quad\land\quad\phi=\frac{1+\sqrt{5}}{2} $$ 
I have tried to prove its convergence with the Root Test and the Ratio Test because of the $n$ exponent but can't manage to do it because of the difference in the fraction.
Can anyone help me? Thank you
 A: Since $$F_{n}=\frac{\phi^{n}-\left(-\phi\right)^{-n}}{\sqrt{5}}$$ we have $$F_{n}\sim\frac{\phi^{n}}{\sqrt{5}}$$ as $n\rightarrow\infty$ so $$\sum_{n\ge1}\frac{1}{F_{n}}\sim\sqrt{5}\sum_{n\ge1}\frac{1}{\phi^{n}}=\frac{\sqrt{5}}{\phi-1}.$$
A: You may prove by induction that for any $n\geq 5$ we have $F_{n+5}\geq 11 F_n$. That is enough to deduce convergence by comparison with a geometric series and further get that:
$$ S = \sum_{n\geq 1}\frac{1}{F_n} =\frac{17}{6}+\sum_{n\geq 5}\frac{1}{F_n} = \frac{17}{6}+\sum_{n=5}^{9}\frac{1}{F_n}+\sum_{n\geq 10}\frac{1}{F_n}\leq \frac{17}{6}+\frac{88913}{185640}+\frac{1}{11}\sum_{n\geq 5}\frac{1}{F_n}$$
such that:
$$ \frac{10}{11}\sum_{n\geq 5}\frac{1}{F_n}\leq \frac{88913}{185640},\qquad S\leq \frac{17}{6}+\frac{11}{10}\cdot \frac{88913}{185640}=\frac{2079281}{618800}. $$
A: For $n \ge 3$, we have
$$\frac{1}{F_n} \le \frac{F_{n-1}}{F_nF_{n-2}}
= \frac{F_{n} - F_{n-2}}{F_nF_{n-2}} = \frac{1}{F_{n-2}} - \frac{1}{F_n}
= \left(\frac{1}{F_{n-2}} + \frac{1}{F_{n-1}}\right) - \left(\frac{1}{F_{n-1}} + \frac{1}{F_n}\right)\\
= \frac{F_{n}}{F_{n-2}F_{n-1}} - \frac{F_{n+1}}{F_{n-1}F_n}
$$
The partial sums is monotonic increasing and bounded from above.
$$\sum_{n=1}^N \frac{1}{F_n} = 2 + \sum_{n=3}^N \frac{1}{F_n}
\le 2 + \frac{F_3}{F_1F_2} - \frac{F_{N+1}}{F_{N-1}F_N} \le 2 + \frac{2}{1\cdot 1} = 4$$
As a result, the series converges.
A: $$\frac{F_{n+1}}{F_n}=\frac{\phi^{n+1}-(-\phi)^{-n-1}}{\phi^n-(-\phi)^{-n}}\ge\phi\frac{1-\phi^{-2n-2}}{1+\phi^{-2n}}.$$
The expression on the right is an increasing function of $n$ that exceeds $1$ as of $n=2$ (and quickly tends to $\phi$).
A: Let $F_n$ be the fibonacci sequence 
We know $F_{2n+2}=F_{2n+1}+F_{2n}\geq 2F_{2n}$
Similarly $F_{2n+1}=F_{2n}+F_{2n-1}\geq 2F_{2n-1}$
So $1/F_2+1/F_4+1/F_6\dots<1/F_2+1/2F_2+1/4F_2\dots=1/F_2(1/2+1/4+1/8\dots)$
$1/F_1+1/F_3+1/F_5\dots<1/F_1+1/2F_1+1/4F_1\dots=1/F_1(1/2+1/4+1/8\dots)$
Now I think it's clearly evident that why the sum of reciprocals of the Fibonacci sequence is convergent, only the definition of the Fibonacci sequence is enough!
A: Suppose
$2
\ge \frac{F_{n+1}}{F_n}
\ge \frac32
$
for $n$ and $n+1$.
Then
$\frac{F_{n+2}}{F_{n+1}}
=\frac{F_{n+1}}{F_{n+1}}+\frac{F_{n}}{F_{n+1}}
=1+\frac{F_{n}}{F_{n+1}}
\ge 1 + \frac12
=\frac32
$
and
$\frac{F_{n+2}}{F_{n+1}}
=\frac{F_{n+1}}{F_{n+1}}+\frac{F_{n}}{F_{n+1}}
=1+\frac{F_{n}}{F_{n+1}}
\lt 1 + 1
=2
$.
A: I think I have an interesting proof for the summability of the reciprocals of the FN. It is based on the Kummer's test and the fact that $F_{n-1}\geq  2 F_{n}$, $n\geq 2$. 
First note that $(1/F_n)$ is summable, if and only if, there exists a sequence $(u_n)$ of positive numbers such that 
$$u_n F_{n+1}-u_{n+1}F_{n}\geq F_{n},$$
for all $n$ sufficiently large. (This is the Kummer's test applied to the sequence $(1/F_n)$).
Now, from the definition of FN: $F_{n+1}-F_{n}=F_{n-1}$, for all $n\geq 2$.
That is, taking $u_n = 2$, for all $n\geq 1$, we have that
$2 F_{n+1}- 2 F_{n}= 2 F_{n-1}\geq F_{n}$, for all $n\geq 2$.
As so, by the Kummer's test, the sequence $(1/F_n)$ is summable. 
