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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.

Somos
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Matteo
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    What would happen if we mirrored the proofs in this post? – Тyma Gaidash Sep 06 '23 at 13:23
  • @Tyma Gaidash: I don't know, can tou explain? – Matteo Sep 06 '23 at 18:59
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    @ТymaGaidash٠ If I'm not mistaken, when shifting the index by 1 there is a sign change in that crucial identity, so we would probably get an identity about artanh instead of arctan. – Trebor Sep 08 '23 at 05:51
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    I did the computation myself and the last few digits of your calculation disagreed with mine. Here is the constant I found, given to 215 decimal places, checked with the mpmath Python library and Mathematica: 1.30850 28221 75405 19611 19578 86489 00316 25205 25052 94799 80042 15302 15949 90724 21084 65827 04822 07964 13690 57342 23668 44455 03065 19198 89495 10460 89504 66944 28478 35459 20105 02988 61244 93429 95108 35884 48077 27227 84623 60198 92057 83514 60081 – ho boon suan Sep 08 '23 at 07:41
  • @hoboonsuan: it may be a machine precision problem. I will update the question. – Matteo Sep 08 '23 at 10:38
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    I note that $\sum_{n\ge1}\arctan{(-1)^n\over F_{2n}}=\sum_{n\ge1}(-1)^{n+1}\arctan{1\over F_{2n}}=\arctan(-\beta)$, where $\beta=(1-\sqrt5)/2$ (see for example Corollaries 26.2 and 40.3 in Thomas Koshy's book Fibonacci and Lucas Numbers with Applications (2 volumes)); together with the identity $\sum_{n\ge1}\arctan{1\over F_{2n+1}}=\pi/4$, this means that it is an equivalent problem to determine $\sum_{n\ge1}\arctan{1\over F_n}$ or $\sum_{n\ge1}\arctan{1\over F_{4n}}$. – ho boon suan Sep 09 '23 at 08:13
  • By using atan instead of std::atan or atanl you're probably truncating the series terms to double precision. Also, even long double is unlikely to have 50 significant decimal digits. – Daniel Schepler Sep 14 '23 at 17:53

1 Answers1

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}

Jakob
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  • and then? $$\sum_{k=1}^{+\infty} {\arctan\left (\frac{m_k^2-n_k^2}{2 m_k n_k}\right)}$$has a closed form? – Matteo Sep 17 '23 at 05:57
  • That would be nice but as you have asked .... On the infinite series ... a simple closed form is not very likely. You can improve the series combining more terms in one e.g. $..{}k and ..{}{k+1}$. – Jakob Sep 17 '23 at 08:03