Fibonacci sequences and related series 
Let $\{a_n\}$ be a sequence such that $a_1=a_2=1$ and $a_{n+1}=a_n+a_{n-1}$ for $n\geq 2$. Prove that $\displaystyle \sum_{n=1}^\infty \frac{1}{a_n}$ converges.

My work:
Let $b_n=\frac{1}{a_n}$. Then I proved that $|\frac{b_{n+1}}{b_n}|<1$. I was going to use ratio test. But how can I show that limsup $|\frac{b_{n+1}}{b_n}|\neq1$? Can anyone please help me?
 A: Can you assume the golden ratio is the limit of the ratio of fibonacci numbers?
$$\dfrac{b_{n+1}}{b_n} = \dfrac{1}{\dfrac{a_{n+1}}{a_n}} \to \dfrac{1}{\phi}$$
A: With somewhat less artillery: from $a_{n+1} = a_n + a_{n-1}$ we have 
$a_{n+1}/a_n = 1 + a_{n-1}/a_n$, so if $r_n = a_{n+1}/a_n$ that says
$r_n = 1 + 1/r_{n-1}$.
From this and $r_1 = 1$, $r_2 = 2$ it's not hard to see that $r_n \ge 3/2$ for
$n \ge 2$, which is exactly the sort of bound you need.
A: Bazooka-to-kill-a-fly approach:
Note that $a_n=f_n$, where $f_n$ is the $n$th Fibonacci number.  Thus, $a_n=[\phi^n/\sqrt 5]$, where $\phi = \frac{1+\sqrt 5}{2}$ and $[\cdot]$ is the nearest integer function.
Then:
\begin{align}
\sum_{n=1}^\infty \frac{1}{a_n} &=\sum_{n=1}^\infty \frac{1}{[\phi^n/\sqrt5]}\\
&\leq \sum_{n=1}^\infty \frac{\sqrt 5}{\phi^n} \\
&= \frac{\sqrt 5}{1-\frac{1}{\phi}}
\end{align}
A: A cute answer starts with Cassini's identity
$$
F_{n+1}F_{n-1} - F_n^2 = (-1)^n
$$
which is easy to prove by induction.
Consider the ratio $\frac{F_{n+1}}{F_n}$.  If we show this is always strictly greater than one for $n>1$, then limsup $|\frac{b_{n+1}}{b_n}| : n > 1$ is less than $1$.
For $n$ odd, $\frac{F_{n+1}}{F_n} > \frac{F_{n-1}}{F_{n-2}}$ because
$$
\frac{F_{n+1}}{F_n} = \frac{F_n}{F_{n-1}}-\frac{1}{F_{n-1}} = \frac{F_{n-1}}{F_{n-2}} - \frac{1}{F_{n-1}} + \frac{1}{f_{n-2}} >  \frac{F_{n-1}}{F_{n-2}}-\frac{1}{F_{n-2}} + \frac{1}{f_{n-2}} = \frac{F_{n-1}}{F_{n-2}}$$
So the smallest value of $\frac{F_{n+1}}{F_n}$ for odd $n$ greater than 1 occurs at $n=3$, and is $\frac{3}{2}$.
Further, for all even $n$,  $\frac{F_{n+1}}{F_n} >  \frac{F_{n}}{F_{n-1}}$ again by Cassini's identity.
So liminf $\frac{F_{n+1}}{F_n} = \frac{3}{2}$ and 
$|\frac{b_{n+1}}{b_n}| > \frac{2}{3}$  for all $n > 1$.
