I want to check if $$\left\lfloor \left( \sum_{k=n}^{2n}{\frac{1}{F_{2k}}} \right)^{-1} \right\rfloor =F_{2n-1}~~(n\ge 3) \tag{$*$}$$ where $\lfloor x \rfloor$ is th floor function.

The Fibonacci sequence is defined by $F_1=1$, $F_2=1$, $F_{n+1}=F_n+F_{n-1}~(n\ge 2)$. Then we can get $$F_n=\dfrac{\alpha^n-\beta^n}{\sqrt{5}}$$ where $\alpha=\dfrac{1+\sqrt{5}}{2}$ and $\beta=\dfrac{1-\sqrt{5}}{2}.$

The following are some of my attempts:

For some example:

$n=3$, the left hand is $5$, the right hand is $5.$

$n=4$, the left hand is $13$, the right hand is $13.$


$n=15$, the left hand is $514229$, the right hand is $514229.$

It is all true. But as $n$ increases, the order of magnitude grows very rapidly.

I ask one of my good friends to use a Python program to check $(*).$ He says it is true for $n\le 35$. When $n=36$, the Python says it is not true, But when $n= 37$, it is true again.

Thus I change one way and I ask my fiend to use a Python program to check $$\left( \sum_{k=n}^{2n}{\frac{1}{F_{2k}}} \right)^{-1} =F_{2n-1}~~(n\ge 3) \tag{$**$}.$$

Then the program shows it is true at least for $31\le n\le 51.$

But as you see, the left hand of $(**)$ is a decimal and the right hand of $(**)$ is an integer.

So I do not know if it is because the order of magnitude on the left hand of $(*)$ is growing very fast, $(*)$ becomes not true due to some computer shortcomings.

Finally I wonder if $(*)$ is true or false? Any help and references are greatly appreciated.


I have also posted it on MSE as On the finite sum of reciprocal Fibonacci sequences

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    $\begingroup$ I've checked with PARI/GP, the equality is true for $1\le n\le500$. $\endgroup$ Aug 5, 2023 at 9:48
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    $\begingroup$ This seems pretty similar: artofproblemsolving.com/community/c6h1355788p7417584 $\endgroup$ Aug 5, 2023 at 10:19
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    $\begingroup$ OP cross-posted to MathSE: math.stackexchange.com/q/4747937 $\endgroup$ Aug 5, 2023 at 10:32
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    $\begingroup$ See also A. Y. Wang and P. Wen, “On the partial finite sums of the reciprocals of the Fibonacci numbers”, J. Inequal. Appl. (2015), Article 73. doi.org/10.1186/s13660-015-0595-6 $\endgroup$ Aug 5, 2023 at 10:35
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    $\begingroup$ Generalizing the above, numerical evidence suggests that for all $m\ge1$ and sufficiently large $n$ we have the identity $$\Big\lfloor\Big(\sum_{k=n}^{2n}{1\over F_{mk}}\Big)^{-1}\Big\rfloor=\big((-1)^mF_m/\phi+(-1)^{m-1}F_{m-1}+1\big)F_{mn},$$ where $\phi=(1+\sqrt5)/2$ is the golden ratio. $\endgroup$ Aug 5, 2023 at 13:42

3 Answers 3


We need to prove, equivalently
$$\frac1{F_{2n-1}+1}<\sum_{k=n}^{2n}\frac1{F_{2k}}\le \frac1{F_{2n-1}}, $$ that is, by the above expression for $F_{k}$, since $\beta=-\alpha^{-1}$, we need to check the double inequality $$\frac1{\alpha^{2n-1}+\alpha^{-2n+1}+\sqrt5}<\sum_{k=n}^{2n}\frac1{\alpha^{2k}-\alpha^{-2k}}\le \frac1{\alpha^{2n-1}+\alpha^{-2n+1}}. $$

To do so we bound below and above the middle sum the obvious way

$$\sum_{k=n}^{2n} \frac1{\alpha^{2k}} <\sum_{k=n}^{2n}\frac1{\alpha^{2k}-\alpha^{-2k}} \le \frac1{1-\alpha^{-4n}} \sum_{k=n}^{2n}\frac1{\alpha^{2k}}.$$

By computation $\displaystyle\sum_{k=n}^{2n} \frac1{\alpha^{2k}}= \alpha^{-2n+1}-\alpha^{-4n-1}$, so everything follows from

$$\frac1{\alpha^{2n-1}+\alpha^{-2n+1}+\sqrt5}\le \alpha^{-2n+1}-\alpha^{-4n-1} $$

and $$\frac{\alpha^{-2n+1}-\alpha^{-4n-1}}{1-\alpha^{-4n}} \le \frac1{\alpha^{2n-1}+\alpha^{-2n+1}} $$ both easily checked (the former is true for any $n\ge0$, the latter for $n\ge3$).

  • $\begingroup$ Choose $n=1$, the left hand is $0$ and the right hand is $1$. $n=2$, the left hand is $1$ and the right hand is $2$. But your proof seems it is right for $n\ge 1.$ $\endgroup$
    – fusheng
    Aug 6, 2023 at 14:22
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    $\begingroup$ @ Pietro Majer Sorry, I ignore the latter is for $n\ge 3$. Thanks! $\endgroup$
    – fusheng
    Aug 6, 2023 at 14:28

Let me sketch a proof that this identity holds for big enough $n$. In fact, we can show that

$$\left(\sum_{k = n}^{2n} \frac{1}{F_{2k}}\right)^{-1} = F_{2n-1} + \frac{1}{\varphi \sqrt{5}} + o(1),$$ where $\varphi = \frac{1+\sqrt{5}}{2}$ is of course a golden ratio. Since $0 < \frac{1}{\varphi\sqrt{5}} < 1$ we get the result.

The key idea is to use Binet's formula $F_n = \frac{\varphi^n - \varphi^{-n}}{\sqrt{5}}$ and that if we drop the second term and just use the approximation $F_n \approx \frac{\varphi^n}{\sqrt{5}}$ then the relative error we will get in the left-hand side of $(*)$ is at most $\frac{\varphi^{-2n}}{\varphi^{2n}} = \varphi^{-4n}$ (because this is the biggest possible relative error of each term), which gives us in the end an absolute error of $O(\varphi^{-2n})$ which is very much $o(1)$.

So, we can safely use this approximation and with it our sum becomes just a geometric progression, which we fortunately know how to compute, so I will just present the result (I hope I didn't mess it up...): $$\sqrt{5} \varphi^{-2n} \frac{1-\varphi^{-2n-2}}{1-\varphi^{-2}}.$$

When we invert it and use that $1-\varphi^{-2} = \varphi^{-1}$, we get $$\frac{\varphi^{2n-1}}{\sqrt{5}} \frac{1}{1-\varphi^{-2n-2}} = \frac{\varphi^{2n-1}}{\sqrt{5}} + \frac{1}{\varphi \sqrt{5}} + o(1).$$

Now, it remains to recall that $F_{2n-1} = \frac{\varphi^{2n-1}}{\sqrt{5}} + o(1)$ to get the desired conclusion.

Note that this proof very much relied on the fact that the summation stops at exactly $2n$ (or at least $2n + O(1)$), if it were much less then we would greately overshoot, and if it were much larger then the error term analysis would've been much harder.

I leave it to someone more masochistic to make this effective and cover all $n$.


For further discussion on such sums (the case of infinite series) and generalizations can be found in this paper starting on page 12 and references therein: https://users.math.msu.edu/users/bsagan/Papers/Old/gfp.pdf


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