<s>[INST] Closed form solution for $\sum_{n=1}^\infty\frac{1}{1+\frac{n^2}{1+\frac{1}{\stackrel{\ddots}{1+\frac{1}{1+n^2}}}}}$. 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).
 [/INST] This is not a full answer but a partial result.
You can prove by induction that the described fraction of yours with $k$ horizontal lines is equal to $\frac{F_{k-1}n^2+F_{k}}{F_{k-2}n^4+2F_{k-1}n^2+F_k}$ with $k\ge3$. Perhaps with partial fractioning, you can compute the series using the well known result:
$$
\sum_{n=1}^{\infty}\frac{1}{n^2+z^2}=\frac{\pi z\coth(\pi z)-1}{2z^2}
$$
As a partial result you can compute the limit directly:
$$
\lim_{k\to\infty}\frac{F_{k-1}n^2+F_{k}}{F_{k-2}n^4+2F_{k-1}n^2+F_k}=\lim_{k\to\infty}\frac{F_{k-1}}{F_{k-2}}\frac{n^2+\frac{F_{k}}{F_{k-1}}}{n^4+2\frac{F_{k-1}}{F_{k-2}}n^2+\frac{F_k}{F_{k-2}}}=\varphi\frac{n^2+\varphi}{n^4+2\varphi n^2+\varphi^2}=\frac{\varphi}{n^2+\varphi}
$$
And therefore:
$$
\lim_{k\to\infty}\sum_{n=1}^\infty\frac{1}{1+\frac{n^2}{1+\frac{1}{\stackrel{\ddots}{1+\frac{1}{1+n^2}}}}}=\sum_{n=1}^{\infty}\frac{\varphi}{n^2+\varphi}=\varphi\frac{\pi \sqrt{\varphi}\coth(\pi \sqrt{\varphi})-1}{2\varphi}=\frac{\pi \sqrt{\varphi}\coth(\pi \sqrt{\varphi})-1}{2}
$$
I hope this is helpful.
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