If $\{a_n\}$ are the Fibonacci numbers, then does $\sum_{n=0}^{\infty} \frac{1}{a_n}$ converge? Examine the convergence of the series
$$\sum_{n=0}^{\infty} \frac{1}{a_n}$$
Where $a_0 = 1, a_1 = 2$, and $a_n = a_{n-1} + a_{n-2}$ for $n \ge 2.$
I'm a little confused here. My professor stated that we are not allowed to use any recurrence relation or $\sqrt 5$, but I'm not sure I even see how either of those would help anyway. What is the significance of $\sqrt 5 $? I could find the partial sum but I'm not sure that would help. Any tips are appreciated.
 A: Since $\{a_n\}_{n\geq 0}$ is increasing and $a_{n+2}=a_{n+1}+a_n$ we have, for any $N\geq 2$,
$$ \sum_{n=0}^{N}\frac{1}{a_n} = \frac{3}{2}+\sum_{n=0}^{N-2}\frac{1}{a_n+a_{n+1}} \leq \frac{3}{2}+\frac{1}{2}\sum_{n=0}^{N}\frac{1}{a_n}$$
and by rearranging
$$ \sum_{n=0}^{N}\frac{1}{a_n}\leq 3 \qquad \forall N\geq 2 $$
ensuring convergence. One does not really need any observation made in the comments.
Similarly, by denoting $S=\sum_{n\geq 0}\frac{1}{a_n}=\sum_{n\geq 2}\frac{1}{F_n}$ we have
$$ S = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\sum_{n\geq 4}\frac{1}{a_n}\leq \frac{61}{30}+\frac{1}{2}\left(S-\frac{3}{2}\right) $$
and $S\leq\frac{77}{30}$ is a decent bound. Actually $S\leq 2.36$.

Slight generalization: if $a_0>0$ and $a_{n+k}$ is a linear combination of $a_{n},a_{n+1},\ldots,a_{n+k-1}$ with positive coefficients, whose sum exceeds one, then $\sum_{n\geq 0}\frac{1}{a_n}$ is convergent.

If we combine the original elementary approach with the fact that $\frac{F_{n+2}}{F_n}$ rapidly converges to $\varphi^2=\varphi+1$ we get
$$ 2S-1=\sum_{n\geq 0}\frac{1}{a_n}+\frac{1}{a_{n+1}}=\sum_{n\geq 0}\frac{a_{n+2}}{a_n a_{n+1}}=\frac{2647}{780}+\sum_{n\geq 5}\frac{a_{n+2}}{a_n a_{n+1}}\approx \frac{2647}{780}+\varphi^2\sum_{n\geq 6}\frac{1}{a_n} $$
and $S\approx \frac{3547+60\sqrt{5}}{1560}$ is correct up to the third figure.
A: The significance of $\sqrt{5}$ is due to the relationship with the closed form of the Fibonacci numbers.

However, just proving convergence is much easier. You can show without too much effort that $a_n$ grows exponentially. A brief induction proof will show that, for example, $a_n \ge 1.1^n$ for all $n$. After this, comparison to a geometric series gives convergence.
A: We shall first show, inductively, that 
$$
a_n\ge \frac{3^n}{2^n}.
$$
Clearly, this is true for $n=0,1$. Assume this in true for $n=0,1, \ldots,k$.
Then
$$
a_{k+1}=a_k+a_{k-1}\ge \frac{3^k}{2^k}+\frac{3^{k-1}}{2^{k-1}}=\frac{3^k+2\cdot3^{k-1}}{2^k}=\frac{10\cdot3^{k-1}}{2^{k+1}}>\frac{3^{k+1}}{2^{k+1}},
$$
which completes the inductive proof.
Now, the series $\sum\frac{1}{a_n}$ is dominated by a decaying geometric series since,
$$
0<\frac{1}{a_n}\le \frac{2^n}{3^n},
$$
and hence Comparison Test provides the convergence of $\sum\frac{1}{a_n}$.
