Inequality with Fibonacci numbers $ \sum \limits_{k=1}^{2n+1} (-1)^{k+1} F_k \: \text{arccot} F_k < \frac{\pi+1-\sqrt{5}}{2}$

Question

Prove that $$ \sum \limits_{k=1}^{2n+1} (-1)^{k+1} F_k \: \text{arccot} F_k < \frac{\pi+1-\sqrt{5}}{2}$$ holds for all $n \in \mathbb{N}$.

(The Fibonacci sequence, defined by the recurrence $F_1 = F_2 = 1$ and $\forall n \in \mathbb{N},$ $F_{n+2} = F_{n+1} + F_n$)

My work. I proved that the sequence $a_n=\sum \limits_{k=1}^{2n+1} (-1)^{k+1} F_k \: \text{arccot} F_k \;$ is increasing. Therefore inequality cannot be proved by induction.

Answer 1

Not a full answer.

Let's group even and odd terms in the sum:

$$S_n=\sum_{k=1}^{2n+1} (-1)^{k+1} F_k \operatorname{arccot} F_k= \\ = \sum_{l=0}^{n-1} (F_{2l+1} \operatorname{arccot} F_{2l+1}-F_{2l+2} \operatorname{arccot} F_{2l+2}) +F_{2n+1} \operatorname{arccot} F_{2n+1}$$

It's easy to prove that $F_{2n+1} \operatorname{arccot} F_{2n+1} < 1$ and:

$$\lim_{n \to \infty} F_{2n+1} \operatorname{arccot} F_{2n+1}=1$$

So we only need to find the limit for the rest of the sum.

Denote:

$$P_n=\sum_{l=0}^{n-1} (F_{2l+1} \operatorname{arccot} F_{2l+1}-F_{2l+2} \operatorname{arccot} F_{2l+2})$$

The first terms is $0$, it's easy to check. Also, this sum converges to a single limit, so we can denote:

$$P=\sum_{l=1}^\infty \left(F_{2l+1} \arctan \frac{1}{ F_{2l+1} } -F_{2l+2} \arctan \frac{1}{ F_{2l+2} } \right)$$

We need to prove that ($\varphi$ - Golden ratio):

$$P= \frac{\pi}{2}-\varphi$$

In this post it is shown how to prove the following identity:

$$\frac{\pi}{2} = 2 \sum_{l=1}^\infty \arctan \frac{1}{ F_{2l+1} }$$

So we are left to prove that:

$$Q=\sum_{l=1}^\infty \left(F_{2l+2} \arctan \frac{1}{ F_{2l+2}}- (F_{2l+1}-2) \arctan \frac{1}{ F_{2l+1} } \right)= \varphi$$

Numerically this works. The proof eludes me.

I've spent a few hours trying different methods, but haven't had much luck so far.

Some useful identities:

$$\arctan \frac{1}{a} = \int_0^1 \frac{a dt}{t^2+a^2}$$

$$\arctan \frac{1}{a} = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} \frac{1}{a^{2k+1}}$$

$$F_n= \frac{1}{\sqrt{5}} \left(\varphi^n -\frac{(-1)^n}{\varphi^n} \right)$$

Notice that we can write the odd terms as:

$$F_{2l+1}=\frac{2}{\sqrt{5}} \cosh ((2l+1) \alpha)$$

$$F_{2l+2}=\frac{2}{\sqrt{5}} \sinh ((2l+2) \alpha)$$

Where $$\alpha= \log \varphi$$

Let's not forget the special property of the golden ratio:

$$\varphi-1=\frac{1}{\varphi}$$

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There's also a more general arctangent identity listed on this page:

$$\arctan \frac{1}{ F_{2l} }= \sum_{j=l}^\infty \arctan \frac{1}{ F_{2j+1} }$$

Answer 2

Because of the size, I'm writing this attempt as a separate answer.

Still no closer to the proof of the closed form, but I decided to try Euler-Maclaurin, since the integrals can be evaluated explicitly.

We change the sign for $P$ from my other answer, so it will be positive, and work with it:

$$P=\sum_{l=1}^\infty \left(F_{2l+2} \arctan \frac{1}{ F_{2l+2} } -F_{2l+1} \arctan \frac{1}{ F_{2l+1} } \right)=\varphi- \frac{\pi}{2}= \\ = 0.047237661954998228973 \ldots$$

Using the integral representation of arctangent and the hyperbolic form of Fibonacci numbers, we have:

$$P= \int_0^1 \sum_{l=1}^\infty \left( \frac{F_{2l+21}^2}{ t^2+F_{2l+2}^2 } - \frac{F_{2l+1}^2}{ t^2+F_{2l+1}^2 } \right) dt$$

Simple rearrangement leads to the sum of two convergent series:

$$P= \int_0^1 \sum_{l=1}^\infty \left( \frac{t^2}{ t^2+F_{2l+1}^2 } - \frac{t^2}{ t^2+F_{2l+2}^2 } \right) dt$$

Or, explicitly:

$$P= \int_0^1 \sum_{l=1}^\infty \left( \frac{ \frac54 t^2}{ \frac54 t^2+\cosh^2 (2l+1) \alpha } - \frac{\frac54 t^2}{\frac54 t^2+\sinh^2 (2l+2) \alpha } \right) dt$$

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Let's denote $\frac54 t^2=y^2$ and consider the series:

$$S_1(y)=\sum_{l=1}^\infty \frac{ 1}{y^2+\cosh^2 (2l+1) \alpha }$$

$$S_2(y)=\sum_{l=1}^\infty \frac{ 1}{y^2+\sinh^2 (2l+2) \alpha }$$

I doubt the series have a general closed form, so let's try our hand at Euler-Maclaurin:

$$S_1(y) = \int_1^\infty \frac{ dx}{y^2+\cosh^2 (2x+1) \alpha }+ \\ +\frac{1}{2} \frac{ 1}{y^2+\cosh^2 3\alpha }- \sum_{k=1}^\infty \frac{B_{2k}}{(2k)!} \frac{\partial^{2k-1}}{\partial x^{2k-1}} \left(\frac{ 1}{y^2+\cosh^2 (2x+1) \alpha } \right) \bigg|_{x=1}$$

$$S_2(y) = \int_1^\infty \frac{ dx}{y^2+\sinh^2 (2x+2) \alpha }+ \\ +\frac{1}{2} \frac{ 1}{y^2+\sinh^2 4\alpha }- \sum_{k=1}^\infty \frac{B_{2k}}{(2k)!} \frac{\partial^{2k-1}}{\partial x^{2k-1}} \left(\frac{ 1}{y^2+\sinh^2 (2x+2) \alpha } \right) \bigg|_{x=1}$$

Where $B_{2k}$ are Bernoulli numbers. Here I assumed that all the derivatives at $x \to \infty$ give exactly $0$, which is a reasonable assumption. Even though the series is asymptotic, I suspect it converges in this case.

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Both the double integrals can be found exactly, but I won't write the full solution here since it's too long, I'll just provide the values from Mathematica:

$$ \int_0^1 \int_1^\infty \frac{5/4 t^2 dx dt}{5/4 t^2+\cosh^2 (2x+1) \alpha }= \\ =\frac{\sqrt{5} \left(\log 8-3 \log \left(35-15 \sqrt{5}\right)+4 \arctan \left(\frac{4}{3}\right)\right)-10}{20 \alpha}=0.0425812 \ldots$$

$$ \int_0^1 \int_1^\infty \frac{5/4 t^2 dx dt}{5/4 t^2+\sinh^2 (2x+2) \alpha }= \\ = \frac{\sqrt{5}\left(\log 4- \log \left(15-5\sqrt{5}\right)+14 \arctan \left(\frac{1}{3}\right)\right)-10}{20 \alpha}=0.0182396 \ldots$$

Now for the second terms of the E-M expansion:

$$\frac{1}{2} \int_0^1 \frac{5/4 t^2 dt}{5/4 t^2+\cosh^2 3 \alpha }=\frac{1}{2} - \tan^{-1} \frac{1}{2}$$

$$\frac{1}{2} \int_0^1 \frac{5/4 t^2 dt}{5/4 t^2+\sinh^2 4 \alpha }=\frac{1}{2} - \frac{3}{2}\tan^{-1} \frac{1}{3}$$

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Let's see what we get so far by adding all four terms:

$$P_0= \frac{ -2\log 5-\log \left(123-55 \sqrt{5}\right)+4 \arctan \left(\frac{4}{3}\right)-14 \arctan \left(\frac{1}{3}\right))}{4 \sqrt{5} \alpha}+\frac{3}{2}\tan^{-1} \frac{1}{3}-\tan^{-1} \frac{1}{2} = \\ = 0.043319806992170091958 \ldots$$

This doesn't simplify much (at least Mathematica can't do it), and the numerical value is not that close to the desired answer.

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Considering that $B_2=1/6$, taking the first derivatives and evaluating the integrals w.r.t. $t$ gives us an additional correction:

$$\Delta_1 P = \frac{ \left(1+8 \arctan \left(\frac{1}{2}\right)-14 \arctan \left(\frac{1}{3}\right)\right) \alpha}{12 \sqrt{5}}$$

$$P_0+\Delta_1 P=0.046990354355256824367$$

Which is closer to the answer. Taking more derivatives, we get various more complicated terms with increasing odd powers of $\alpha$.

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This was fun, but I don't see any reason to continue. Let's just say, this method might be interesting for more general series, while this case with Fibonacci numbers obviously has some simple trick that can be used to prove the closed form.