# Sum $\frac{1}{1\times2}+\frac{1}{1\times3}+\frac{1}{2\times5}+\frac{1}{3\times8}+\cdots$

If $f_n$ is the Fibonacci series, with $1,1,2,3,5,8,\ldots$ prove that

$$\sum_{i=2}^\infty\frac{1}{f_{i-1}\cdot f_{i+1}} = 1$$

So my idea was to try to convert this series into a telescoping sum somehow, because otherwise I can't see how this would be managed.

$$\frac{1}{1\times2}+\frac{1}{1\times3}+\frac{1}{2\times5}+\frac{1}{3\times8}+\cdots$$

I can't see any obvious way to re-write the terms though. I could try sum this using Binet's formula but I am pretty sure that will get out of hand.

What other alternatives do I have here?

Note: If you use any other identity other than $f_n=f_{n-1}+f_{n-2}$, or Binet's formula. kindly link to, or provide , a proof of it.

## 2 Answers

You could note that

$$\frac{1}{f_nf_{n-1}} - \frac{1}{f_nf_{n+1}} = \frac{1}{f_{n-1}f_{n+1}}$$

Which could help with your telescoping sum.

• sorry for the late acceptance. I wasn't online for a few days. – Guy Mar 29 '14 at 13:38

We can use the identity via partial fractions,

$\dfrac{1}{x(a+x)} = \dfrac{1}{a}\left(\dfrac{1}{x} - \dfrac{1}{a+x} \right)$

with $x = f_{i-1}$ and $a = f_{i}$ (so that $a+x = f_{i+1}$) to get

$\dfrac{1}{f_{i-1}f_{i+1}} = \dfrac{1}{f_i}\left(\dfrac{1}{f_{i-1}} - \dfrac{1}{f_{i+1}} \right) = \dfrac{1}{f_i \ f_{i-1}} - \dfrac{1}{f_{i+1} f_{i}}$.

So if $\phi_i:= \frac{1}{f_i \ f_{i-1}}$,

$\dfrac{1}{f_{i-1}f_{i+1}} = \phi_{i} - \phi_{i+1}$ and we can telescope,

$\displaystyle \sum_{i=2}^N \dfrac{1}{f_{i-1}f_{i+1}} = \phi_2 - \phi_{N+1} = 1 - \phi_{N+1}$

To conclude, realise that $f_i\to\infty$, so $\phi_N\to0$, and so

$\displaystyle \sum_{i=2}^∞ \dfrac{1}{f_{i-1}f_{i+1}} = 1 - \lim_{N→∞}\phi_{N+1} = 1 - 0$.