<s>[INST] A summation of a series based on the Fibonacci sequence. A sequence is defined as follows: $$a_n=a_{n-1}+a_{n-2}\  \forall \ n\geq3\  ,\ n\in Z. $$
If $a_1=a$ and $a_2=b$, find $$S=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}\frac{1}{a_{2i-1}a_{2i+1}}$$ in terms of a and b.

My Approach:
I first found the n-th term as $a_n=F_{n-1}a_2+F_{n-2}a_1$ where $F_i$ is the i-th term of the well-known Fibonacci sequence $1,1,2,3,5,8,13...$
I also noticed that by setting $a=b$ the resulting sum comes to $\frac{1}{a^{2}}\left(\frac{1}{F_1F_3}+\frac{1}{F_3F_5}+\frac{1}{F_5F_7}+...\right)$. And by this theorem,(https://link.springer.com/article/10.1186/s13662-016-0860-0) 
The final sum should be, $\frac{1}{a^{2}}*\frac{1}{F_2}=\frac{1}{a^{2}}$.
I also tried expanding the i-th term of $S$ but it did not simplify well.
Another result which may be of use is $\frac{1}{a_{2i-1}a_{2i+1}}=\frac{1}{a_{2i}}\left(\frac{1}{a_{2i-1}}-\frac{1}{a_{2i+1}}\right)$

A solution or any hint/insight on how to solve this problem will be much appreciated.
Thank you 
 [/INST] Telescopic summation works
Let a=b=1, for simplicity. Use the property of Fibonacci numbers $F_m$:
$$F_{m+1} F_{m+2} -F_m F_{m+3}=(-1)^m.$$ Here $A_m=F_m.$
Let $m=2(n-1)$, we get
$$A_{2n-1}A_{2n} - A_{2n-2} A_{2n+1}=1.$$
Hence $$\frac{1}{A_{2n-1} A_{2n+1}}= \frac{A_{2n-1}A_{2n} - A_{2n-2} A_{2n+1}}
{A_{2n-1} A_{2n+1}}=\frac{A_{2n}}{A_{2n+1}}-\frac{A_{2n-2}}{A_{2n-1}}=B_n-B_{n-1}$$
Telescopic summation:
$$S=\sum_{k=1}^{n} \frac{1}{A_{2n-1} A_{2n+1}}=[(B_1-B_0)+(B_2-B_1)+(B_3-B_2)+...+(B_{n}-B_{n-1}]$$
$$\implies S=B_n-B_0=\frac{A_{2n}}{A_{2n+1}}-0=\frac{A_{2n}}{A_{2n+1}},$$
as $B_0=\frac{F_0}{f_2}=0$
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