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I am currently going through exercises regarding convergence/divergence.

For my previous question I used the ratio test, and managed to get through it all okay (I think). I proved that: $$\sum_{n=1}^{\infty} \frac{n!}{n^n}$$ converges, and now I have to show whether or not an inverse Fibonacci sum converges/diverges and I'm not sure what method to use. What is the best way to tackle this problem?

$$\sum_{n=1}^{\infty} \frac{1}{f_nf_{n+2}}$$ Where $f_n$ is the Fibonacci sequence, $f_n = f_{n-1} + f_{n-2}$ with initial terms $f_1 = f_2 = 1$

I don't believe it's similar to how I completed $\sum_{n=1}^{\infty} \frac{n!}{n^n}$ but let me know if I'm wrong. Based on looking at fairly similar questions on this website I have started trying to use proof by contradiction.

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    $\begingroup$ Looks like Binet's formula could be helfpful... $\endgroup$ – user378947 Nov 19 '16 at 18:53
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Notice $$\frac{1}{f_n f_{n+2}} = \frac{f_{n+1}}{f_nf_{n+1}f_{n+2}} =\frac{f_{n+2}-f_{n}}{f_n f_{n+1} f_{n+2}} = \frac{1}{f_nf_{n+1}}-\frac{1}{f_{n+1}f_{n+2}}$$ We are dealing with a telescoping sum and

$$\sum_{n=1}^\infty \frac{1}{f_nf_{n+2}} = \lim_{p\to\infty} \sum_{n=1}^p \frac{1}{f_nf_{n+2}} = \lim_{p\to\infty}\left(\frac{1}{f_1f_2} - \frac{1}{f_{p+1}f_{p+2}}\right) = \frac{1}{f_1 f_2} = 1$$

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    $\begingroup$ Love your answer!! ^-^ Kudos!! $\endgroup$ – user378947 Nov 19 '16 at 19:09
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    $\begingroup$ I think this answer could be improved by making explicit why the sum converges, since this is mainly a question about convergence. Not all telescoping sums converge; the fact that the terms approach zero is crucial. For many readers this implicit fact might be trivial, but probably not for everyone. $\endgroup$ – Pedro A Nov 20 '16 at 15:52
  • $\begingroup$ Thanks a lot 3 years later 💪 $\endgroup$ – Eden Hazard May 31 at 0:52
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Another approach. The $n$th Fibonacci number is about $\varphi^n$, where $\varphi = (1+\sqrt{5})/2$ is the golden mean. Then your sum behaves like $\Sigma (1/\varphi^{2n})$. It's easy to show that converges. Of course you don't get the value of what it converges to, as in @achillehui 's nice answer.

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  • $\begingroup$ The question doesn't state that I have to find the value. Even though I made @achillehui 's response the answer, I feel you're approach is very 'clean' and outside the box. So useful for other Fibonacci questions. Awesome. $\endgroup$ – B Taylor Nov 19 '16 at 19:12
  • $\begingroup$ @BTaylor, I do not want to take away anything from Ethan's answer, but it is not “outside the box“. Indeed I just up voted it because it is the first approach that one would think of, and it is always good to know a standard way of doing something, even if it turns out that for a particular problem there is a more clever way. $\endgroup$ – Carsten S Nov 20 '16 at 14:39
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Since it is easy to see that $f_n\geq n$ for $n\geq5$ you can bound the (positive) summand $\frac1{f_nf_{n+2}}<\frac1{n^2}$ for $n\geq5$; therefore since $\sum_{n\geq1}\frac1{n^2}$ converges, so does $\sum_{n\geq1}\frac1{f_nf_{n+2}}$.

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