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I am trying to evaluate an infinite series that involves Fibonacci numbers. By adding the first thousands of terms, I get an approximation of the sum: $S = 5.3598856662431775531720113029189271796889051337319684864955538153...$.

I guess that $S$ can be written exactly as $a+b\sqrt{5}$, where $a, b \in \mathbb{Q}$. Suppose for the moment that it is true. To find $a$ and $b$, I wonder if there are any methods similar to finding convergents of continued fractions. Of course, if we suppose $a=0$, then we can find $b \in \mathbb{Q}$ that approximates $\frac{S}{\sqrt{5}}$ as well as we wish. But here I am allowing one more free variable in the hope that the approximation is in fact exact, i.e. $S = a+b\sqrt{5}$.

Edit: To @Lubin, and in case anyone is wondering, $$S = \sum_{k=2}^\infty \frac{F_{k+2}}{F_{k+1}F_{k-1}},$$ where $F_k$ is the $k$th Fibonacci number, with $F_0=0, F_1=1, F_2=1, ...$.

p.s. I am asking this question as a general approximation question. As suggested by @Bonnaduck (and I've also given a proof in the comments), my guess above is very likely to be false because $S$ involves the reciprocal Fibonacci constant.

The main problem is better phrased as: Given a real number $x$ in decimal form (we have a way to get as many digits as we need), find $a, b \in \mathbb{Q}$ with small denominators such that $a+b\sqrt{5}$ is a good approximation of $x$, similar to the traditional convergents method. Moreover, if $x$ is indeed of the form $a+b\sqrt{5}$, then this method should produce such $a$ and $b$ within finite steps.

Joseph
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  • Sorry, but what was $S$? – Lubin Dec 19 '21 at 21:13
  • Without knowing the series, there really isn’t much way to do this, and the series being in terms of Fibonacci numbers doesn’t make it more likely to be possible, except for a very narrow sort of series. So, if for some reason you really can’t tell us the series, not sure what we can do. – Thomas Andrews Dec 19 '21 at 21:17
  • There could be an algorithm for finding $\frac{a}{c} + \frac{b}{d} \sqrt{5}$ close to $S$ where $a,b,c,d$ are "small" integers. But I don't see a clever way to implement that, beyond mostly brute force. – aschepler Dec 19 '21 at 21:21
  • The convergents of $S$ may betray $a,,b$. – J.G. Dec 19 '21 at 21:43
  • @J.G. What do you mean? – Joseph Dec 19 '21 at 21:48
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    Expanded hint: roots of quadratics with integers coefficients have recurring integers in the continued fraction expansion (hopefully with a short period). – J.G. Dec 19 '21 at 22:04
  • Your constant seems to be $2+\mathcal F$, where $\mathcal F$ is the Reciprocal Fibonacci Constant. – Bonnaduck Dec 19 '21 at 22:11
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    @Bonnaduck Thanks! I now realized that we can prove S = 2 + $\mathcal F$ as follows: Break the numerator $F_{k+2}$ into $F_{k+1} + (F_{k+1} - F_{k-1})$. The $F_{k+1}$ term corresponds to the sum of the reciprocals $\frac{1}{F_{k-1}}$, whereas the $(F_{k+1} - F_{k-1})$ term corresponds to a telescoping sum which can easily be evaluated to equal 2. – Joseph Dec 19 '21 at 22:48
  • @J.G. That's a good idea! – Joseph Dec 19 '21 at 23:09
  • See https://en.wikipedia.org/wiki/Integer_relation_algorithm for the general question of checking for integer relations among (presumably) irrational numbers. – Barry Cipra Dec 19 '21 at 23:18
  • Also, if you use the explicit formula for the nth Fibonacci number in your infinite series, you can evaluate it explicitly and systematically by elementary means (without having the luck of observing telescoping), ... via summing geometric series. – paul garrett Dec 19 '21 at 23:22

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