Let $$ \text{S}_k = \sum_{n=1}^\infty\cfrac{1}{1+\cfrac{n^2}{1+\cfrac{1}{\ddots1+\cfrac{1}{1+n^2}}}},\quad\text{$k$ rows in the continued fraction} $$ So for example, the terms of the sum $\text{S}_6$ are $$ \cfrac{1}{1+\cfrac{n^2}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+n^2}}}}}} $$ Using a symbolic computation software (Mathematica), I got the following interesting results: $$ \begin{align} \text{S}_4 &= \frac{\pi}{4}\left(\coth(\pi)+\sqrt{3}\coth(\sqrt{3}\pi)\right)-\frac{1}{2}\\ \text{S}_6 &= \frac{\pi}{4}\left(\sqrt{2}\coth\left(\sqrt{2}\pi\right)+\sqrt{\frac{4}{3}}\coth\left(\sqrt{\frac{4}{3}}\pi\right)\right)-\frac{1}{2}\\ \text{S}_8 &= \frac{\pi}{4}\left(\sqrt{\frac{3}{2}}\coth\left(\sqrt{\frac{3}{2}}\pi\right)+\sqrt{\frac{7}{4}}\coth\left(\sqrt{\frac{7}{4}}\pi\right)\right)-\frac{1}{2}\\ \text{S}_{10} &= \frac{\pi}{4}\left(\sqrt{\frac{5}{3}}\coth\left(\sqrt{\frac{5}{3}}\pi\right)+\sqrt{\frac{11}{7}}\coth\left(\sqrt{\frac{11}{7}}\pi\right)\right)-\frac{1}{2}\\ \text{S}_{12} &= \frac{\pi}{4}\left(\sqrt{\frac{8}{5}}\coth\left(\sqrt{\frac{8}{5}}\pi\right)+\sqrt{\frac{18}{11}}\coth\left(\sqrt{\frac{11}{7}}\pi\right)\right)-\frac{1}{2}\\ \text{S}_{14} &= \frac{\pi}{4}\left(\sqrt{\frac{13}{8}}\coth\left(\sqrt{\frac{13}{8}}\pi\right)+\sqrt{\frac{29}{18}}\coth\left(\sqrt{\frac{29}{18}}\pi\right)\right)-\frac{1}{2}.\\ \end{align} $$

The numbers appearing at the first $\coth$ term are easy to guess: they are the famous Fibonacci numbers.

The numbers at the second $\coth$ term can also be guessed: they appear to be the Lucas numbers. Those are constructed like the Fibonacci numbers but starting with $2,1$ instead of $0,1$.

Hence:

Conjecture:$$\text{S}_{2k}=\frac{\pi}{4}\left(\sqrt{\frac{F_k}{F_{k-1}}}\coth\left(\sqrt{\frac{F_k}{F_{k-1}}}\pi\right)+\sqrt{\frac{L_k}{L_{k-1}}}\coth\left(\sqrt{\frac{L_k}{L_{k-1}}}\pi\right)\right)-\frac{1}{2}$$

I have verified this conjecture for many $k$'s and it always work out perfectly. To me this is quite amazing, but I am not able to verify the conjecture. Can anyone prove it?

Moreover, if true the conjecture implies that

$$\lim_{k\to\infty}\text{S}_{2k}=\frac{\sqrt{\varphi}\pi\coth\left(\sqrt{\varphi}\pi\right)-1}{2}$$

which is also very nice ($\pi$ and $\varphi$ don't meet very often).