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}$ 
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.
 A: 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:
$$\varphi-1=\frac{1}{\varphi}$$

There's also a more general arctangent identity listed on this page:
$$\arctan \frac{1}{ F_{2l} }= \sum_{j=l}^\infty \arctan \frac{1}{ F_{2j+1} }$$
