# On the finite sum of reciprocal Fibonacci sequences

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 $$13.$$

$$\vdots$$

$$n=15$$, 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.

Thanks!

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

• I've checked with PARI/GP, the equality is true for $1\le n\le500$. Aug 5, 2023 at 9:48
• This seems pretty similar: artofproblemsolving.com/community/c6h1355788p7417584 Aug 5, 2023 at 10:19
• OP cross-posted to MathSE: math.stackexchange.com/q/4747937 Aug 5, 2023 at 10:32
• 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 Aug 5, 2023 at 10:35
• 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. Aug 5, 2023 at 13:42

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 $$n\ge3$$).

• 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 $n\ge 1.$ Aug 6, 2023 at 14:22
• @ Pietro Majer Sorry, I ignore the latter is for $n\ge 3$. Thanks! 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