Find the limit of $b_n = \sum_{i=1}^n \frac{1}{a_ia_{i+1}}$ So I got this question in a textbook, it is basically:
Let $a_n$ be a sequence such that
\begin{cases}
a_1=3, a_2=7 \\ a_{n+2}=3a_{n+1}-a_n, \forall n \ge 1
\tag{1}
\end{cases}
and $b_n$ defined dependently on the value of $a_n$ as:
\begin{align}
b_n= \sum_{i=1}^n \frac{1}{a_i a_{i+1}}
\end{align}
Then prove that $b_n$ converges and find $\lim_{n \to \infty} b_n$
My attempts
I definitely believe that there's some way we can prove this recursively, like extracting $a_i$ and $a_{i+1}$ from the products, so we have the classic way to solve Egyptian fractions:
\begin{align}
\frac1{a_ia_{i+1}} = k ( \frac1{a_i}-\frac1{a_{i+1}})
\end{align}
Like that. But I haven't been able to figure it out.
Is there any possible way to do that, or should I try something else? Would $b_n$ has itself a defining formula that is not dependent on $a_i$?
Any help is appreciated!
 A: Let
$$ \alpha = \frac{3-\sqrt{5}}{2}, \qquad \beta = \frac{3+\sqrt{5}}{2} $$
be the zeros of the equation $x^2 - 3x + 1 = 0$. Then by using $\alpha + \beta = 3$ and $\alpha \beta = 1$, we find that
$$ a_{n+1} - \alpha a_n = \alpha (a_{n+2} - \alpha a_{n+1}). $$
From this, we get
$$ \frac{a_2 - \alpha a_1}{a_i a_{i+1}} = \frac{\alpha^{n-1}(a_{i+1} - \alpha a_i)}{a_i a_{i+1}} = \frac{\alpha^{i-1}}{a_i} - \frac{\alpha^i}{a_{i+1}}. $$
So by telescoping,
$$ b_n = \frac{1}{a_2 - \alpha a_1} \left( \frac{1}{a_1} - \frac{\alpha^n}{a_{n+1}} \right). $$
Furthermore it is not hard to check that $(a_i)$ is increasing, hence $a_i \geq 3$ for all $i$. (In fact, we can check that $a_i = \alpha^i + \beta^i$, but this is not necessary in this solution.) Together this and $|\alpha| < 1$, we conclude
$$ \lim_{n\to\infty} b_n = \frac{1}{(a_2 - \alpha a_1) a_1} = \frac{1}{2\sqrt{5}} - \frac{1}{6}. $$
A: Hint: Try to show that $a_{i+1}a_{i-1}-a_{i}^2 = 5$ for all $i \ge 2$. Then, we can write:
$$\dfrac{1}{a_ia_{i+1}} = \dfrac{a_{i+1}a_{i-1}-a_{i}^2}{5a_ia_{i+1}} = \dfrac{1}{5}\left(\dfrac{a_{i-1}}{a_i} - \dfrac{a_i}{a_{i+1}} \right),$$ and the sum will "telescope".
A: First, you have that
$$a_i=\left(\frac{3}{2}-\frac{\sqrt{5}}{2}\right)^i+\left(\frac{3}{2}+\frac{\sqrt{5}}{2}\right)^i$$
So
$$b_i=\frac 1 {a_i\,a_{i+1}}=\frac{2^{2 i+1}}{3\ 2^{2 i+1}+\left(3-\sqrt{5}\right)^{2
   i+1}+\left(3+\sqrt{5}\right)^{2 i+1}}=\frac{1}{L_{2 i}\, L_{2 i+2}}$$
where appear Lucas numbers.
$$b_n=\sum_{i=1}^n b_i$$ generates the sequence
$$\left\{\frac{1}{21},\frac{1}{18},\frac{8}{141},\frac{7}{123},\frac{55}{966},\frac{16}{
   281},\frac{377}{6621},\frac{329}{5778},\frac{2584}{45381},\frac{2255}{39603},\frac{
   17711}{311046},\frac{5152}{90481},\frac{121393}{2131941},\cdots\right\}$$ which converges very fast since
$$\frac {b_{i+1}}{b_i}=\frac{L_{2 i}}{L_{2 i+4}}\to \frac{1}{\phi ^4}$$ Using the last number of the above list, the inverse symbolic calculator already provides
$$\frac{121393}{2131941} \sim   \frac{1}{2 \sqrt{5}}-\frac{1}{6}$$ the difference being $8.86 \times 10^{-12}$.
In fact, there is an explicit formula for the $b_n$; the problem is that it involves the $n^{th}$ derivative of the q-digamma function (this is not the most pleasant function to work with).
