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 
 A: Hint: prove by induction that $\sum_{i=1}^{n}\frac{1}{a_{2i-1}a_{2i+1}} = \frac{F_{2n}}{a(F_{2n-1} a+F_{2n}b)}$
A: 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 $B_0=\frac{F_0}{f_2}=0$
A: You very properly noticed that
$$a_i=a F_{i-2}+b F_{i-1}$$ which makes
$$a_{2i-1}\,a_{2i+1}=\left(a F_{2 i-3}+b F_{2 i-2}\right) \left(a F_{2 i-1}+b F_{2 i}\right)$$
Now, consider
$$S_n= \sum_{i=1}^{n}\frac{1}{a_{2i-1}\,a_{2i+1}}$$ which, using partial fraction decomposition,
$$\frac{1}{a_{2i-1}\,a_{2i+1}}=\frac 1 { \left(F_{2 i} F_{2 i-3}-F_{2 i-2} F_{2 i-1}\right)}\Big[\frac{F_{2 i}}{aF_{2 i-1}+ b F_{2
   i}}-\frac{F_{2 i-2}}{aF_{2 i-3}+b F_{2 i-2}}\Big]$$ that is to say
$$\frac{1}{a_{2i-1}\,a_{2i+1}}=\frac{F_{2 i}}{aF_{2 i-1}+ b F_{2
   i}}-\frac{F_{2 i-2}}{aF_{2 i-3}+b F_{2 i-2}}$$ that is to say
$$S_n= \sum_{i=1}^{n}\frac{1}{a_{2i-1}\,a_{2i+1}}=\sum_{i=1}^{n}\frac{F_{2 i}}{aF_{2 i-1}+ b F_{2
   i}}-\sum_{i=1}^{n}\frac{F_{2 i-2}}{aF_{2 i-3}+b F_{2 i-2}}$$
Just for illustration, the very first terms of this telescoping  sequence are
$$\left\{\frac{1}{a(a+ b)},\frac{3}{a(2 a+3  b)},\frac{8}{a(5 a+8 b)},\frac{21}{a(13
   a+21 b)},\frac{55}{a(34 a+55  b)},\frac{144}{a(89 a+144  b)}\right\}$$ where the coefficients are easy to recognize (the numerator and the coefficient of $b$ are  $F_{2n}$; the coefficient of $a$ is $F_{2n-1}$).
All of that makes
$$S_n=\frac {F_{2n}}{a(a\,F_{2n-1}+b\,F_{2n})}$$ as @jjagmath already showed.
A: Let's try again to reduce the induction proof to a statement about a telescoping sum.
Working backwards, if $b_n=\frac{F_{2n}}{a_1a_{2n+1}}$ then
$$
b_{n}-b_{n-1} = \frac{1}{a_1} \frac{F_{2n}a_{2n-1}-F_{2n-2}a_{2n+1}}{a_{2n-1}a_{2n+1}}.
$$
It remains to simplify the denominator. From three successive recursion equations one can eliminate the even-index terms as
\begin{align}
a_{2n-1}-a_{2n-3}&=a_{2n-2}\\
a_{2n-1}&=a_{2n}-a_{2n-2}\\
a_{2n+1}-a_{2n-1}&=a_{2n}\\
\hline
a_{2n+1}-3a_{2n-1}+a_{2n-3}&=0
\end{align}
This holds equally well for the Fibonacci sequence
$$\begin{align}
\pmatrix{F_{2n}&a_{2n+1}\\F_{2n-2}&a_{2n-1}}
&=\pmatrix{3&-1\\1&0}^{n-1}\pmatrix{F_{2}&a_3\\F_{0}&a_1}
\\[.8em]
F_{2n}a_{2n-1}-F_{2n-2}a_{2n+1}&=\det\pmatrix{1&a_3\\0&a_1}=a_1
\end{align}$$
In total this confirms the term of the series as a telescoping difference, and the thus the circle can be closed,
$$
b_n-b_{n-1}=\frac{1}{a_{2n-1}a_{2n+1}}
\implies 
\sum_{k=1}^n\frac{1}{a_{2k-1}a_{2k+1}}=b_n-b_0=\frac{F_{2n}}{a_1a_{2n+1}}.
$$
