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 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 $n > 1$.
