On the infinite series $\sum_{n=1}^{+\infty}\arctan\left(\frac{1}{F_{2n}}\right)$

Question

Let $F_n$ the $n-$th Fibonacci number, i.e. $F_n=F_{n-1}+F_{n-2}$ with $F_0=0$ and $F_1=1$.

We know that (here): $$\sum_{n=1}^{+\infty}\arctan\left(\frac{1}{F_{2n+1}}\right)=\frac{\pi}{4}$$ but how about the following series: $$\sum_{n=1}^{+\infty}\arctan\left(\frac{1}{F_{2n}}\right)$$

We can use Catalan's identity with $r=2$ and $m=2n$, leading us to: $$F_{2n}^2-F_{2n-2}\cdot F_{2n+2}=(-1)^{2n-2}\cdot F_2^2\implies F_{2n}^2=1+F_{2n-2}\cdot F_{2n+2}$$ So: $$\sum_{n=1}^{+\infty}\arctan\left(\frac{1}{F_{2n}}\right)=\sum_{n=1}^{+\infty}\arctan\left(\frac{1}{\sqrt{1+F_{2n-2}\cdot F_{2n+2}}}\right)$$

How can we go on?

Also, I've calculated the result with the following C++ code and it appears to be $\mathcal{S}=1.30850 28221 75405 19611 19578 86489...$

#include<stdlib.h>
#include<iostream>
#include<math.h>
#include<iomanip>
int main(){

long double fn2, fn1, fn, sum;

fn2 = 0;
fn1 = 1;
sum = 0;

for(int i = 0; i < 5000; i++){
    fn = fn1 + fn2;
    fn2 = fn1;
    fn1 = fn;
     if(i%2==0){
        sum = sum + atan(1/fn);
    }
}

std::cout << setprecision(50);
std::cout << "Result: " << sum << std::endl;
system("pause");
}

With Inverse Symbolic Calculator (via wayback), no result has been found.

Possible linked post here.

Answer 1

A simple way is to adopt the sum of arctangents formula : $$\arctan(a) + \arctan(b) = \arctan\left(\frac{a + b}{1 - a b}\right) $$ to obtain an approximation iterating the Fibonacci sequence. This does not give you a closed form in terms of other functions or constants.

One can rewrite the sum using the Lucas numbers $L_k={2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123,..}$ defining pythagorean triples $(a,b,c)=(m^2-n^2,2mn,m^2+n^2)$ with : \begin{align} m_k = & -\frac{3}{4} +\frac{ (-1)^{4k+1}}{20}+ \frac{ L_{4k+3}}{2}+\frac{ L_{8k+6}}{10} \\ n_k = & -\frac{3}{4} +\frac{ (-1)^{4k+1}}{20} - \frac{ L_{4k+3}}{2}+\frac{ L_{8k+6}}{10} \end{align} which have the properties : \begin{align} m_k^2+n_k^2 -2 m_k n_k= & L_{4k+3}^2 \\ m_k-n_k = & L_{4k+3} \end{align} The Fibonacci sum can then be written in terms of Lucas numbers as follows: \begin{align} \sum_{k=1}^{+\infty} {\arctan\left (\frac{1}{F_{2k}}\right)} = & \frac{\pi}{2}-\arctan\left(\frac{1}{2}\right) +\frac{1}{2} \sum_{k=1}^{+\infty} {\arctan\left (\frac{m_k^2-n_k^2}{2 m_k n_k}\right)} \\ = & 1.3085028221754051961119578864... \end{align}