2
$\begingroup$

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.

$\endgroup$
9
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ sorry for the late acceptance. I wasn't online for a few days. $\endgroup$ – Guy Mar 29 '14 at 13:38
2
$\begingroup$

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

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.