Limit of A Fibonacci Let $P(x)$ be an $n^{th}$ degree-polynomial which is defined below for some odd natural number $n$. And let us denote the set of roots of $P$ by $\{r_1,r_2,r_3,\dots, r_n \}$.
$$\displaystyle P(x) = \sum_{m=0}^n F_mx^m $$
Let ${a_k}$ be a sequence defined by:
$$\displaystyle \frac{1}{a_k-r_ka_{k-1}} = r_{k}r_{k+1}\dots r_n $$
$$\forall k\leq n , k\in N^* \text{, with }a_0=1$$
Also, $\{F_m\} $ is the Fibonacci Sequence $(F_0=F_1=1)$ and the set of roots of $P$ also takes into account multiplicity, so this means that double roots may exist twice.
How do I find:
$$\displaystyle \lim_{n\to \infty} (a_n+1)F_{n+1} $$
 A: Rearranging the given recursive formula:
$$\displaystyle a_k=r_ka_{k-1} + \frac{1}{\prod_{j=k}^nr_j} $$
In general:
$$\displaystyle a_k=B_ka_{k-1} + M_k $$
$$\displaystyle = B_kB_{k-1}a_{k-2} + M_{k-1}B_k +M_k $$
$$\displaystyle = B_kB_{k-1}B_{k-2}a_{k-3} + M_{k-2}B_kB_{k-1} + M_{k-1}B_k +M_k $$
$$= \dots $$
$$\displaystyle a_k = \Bigg(\prod_{p=1}^k B_p \Bigg) \Bigg( a_0 +\sum_{i=1}^k \frac{M_i}{\prod_{t=1}^i B_t} \Bigg) $$
Substitute:$ a_0=1 , B_k = r_k $ and $ M_k = \frac{1}{\prod_{j=k}^nr_j} $ :
$$\displaystyle a_k = \Bigg(\prod_{p=1}^k (r_p) \Bigg) \Bigg( 1 +\sum_{i=1}^k \frac{\frac{1}{\prod_{j=i}^nr_j}}{\prod_{t=1}^i (r_t)} \Bigg) $$
$$\displaystyle = \Bigg(\prod_{p=1}^k r_p \Bigg) \Bigg( 1 +\sum_{i=1}^k \frac{1}{r_i\prod_{j=1}^n r_j} \Bigg) $$
Let $k=n$:
$$\displaystyle a_n= \Bigg(\prod_{p=1}^n r_p \Bigg) \Bigg( 1 +\sum_{i=1}^n \frac{1}{r_i\prod_{j=1}^n r_j} \Bigg) $$
Note that $ \{r_1,r_2,\dots,r_n \}$ are the roots of $P(x) = \sum_{m=0}^n F_mx^m $. 
$$\displaystyle \prod_{j=1}^n r_j = (-1)^n\frac{F_0}{F_n} = - \frac{1}{F_n} \;\;\;\; ( n \textrm{ is odd})$$
$$\displaystyle => a_n = -\frac{1-F_n \sum_{i=1}^n \frac{1}{r_i} }{F_n} $$
Now working on the sum :
$$\displaystyle \sum_{i=1}^n \frac{1}{r_i} $$
$$\displaystyle = \frac{\sum_{i=1}^n \frac{\prod_{j=1}^n r_j}{r_i}}{\prod_{j=1}^n r_j} $$
$$\displaystyle = \frac{\sum_{cyc} r_1r_2\dots r_{n-1} }{\prod_{j=1}^n r_j} $$
$$\displaystyle \sum_{cyc} r_1r_2\dots r_{n-1} = (-1)^{n-1}\frac{F_1}{F_n} = \frac{1}{F_n} $$
$$\displaystyle => \sum_{i=1}^n \frac{1}{r_i} = \frac{\frac{1}{F_n} }{-\frac{1}{F_n} } = -1 $$
Substitute back in $a_n$ :
$$\displaystyle a_n = -\frac{1+F_n}{F_n} $$
Now we do the limit:
$$\displaystyle \lim_{n\to \infty} (a_n+1)F_{n+1} $$
$$\displaystyle = - \lim_{n\to \infty} \frac{F_{n+1}}{F_n} $$
$$\displaystyle = -\phi $$
