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.

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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>) [![enter image description here](https://i.stack.imgur.com/MUILw.png)](https://i.stack.imgur.com/MUILw.png)

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 $\frac{1}{a\_{2i-1}a\_{2i+1}}=\frac{1}{a\_{2i}}\left(\frac{1}{a\_{2i-1}}-\frac{1}{a\_{2i+1}}\right)$

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