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).
 A: 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.
A: This answer contains computer-assisted algebra bashing. Pencil wielding mathematicians be warned.
Let
$$f_k(n)=\frac{1}{1+\frac{n^2}{1+\frac{1}{\frac{^\ddots_1}{1+n^2}}}}$$
where there are $2k$ horizontal fraction bars. Then, we have
$$f_2(n)=\frac{1}{1+\frac{n^2}{1+\frac{1}{1+\frac{1}{1+n^2}}}}=\frac{\frac{1}2}{n^2+1}+\frac{\frac{3}2}{n^2+3}$$
$$f_{k+1}(n)=\frac{1}{1+\frac{n^2}{1+\frac{1}{1+\frac{\frac{1}{f_k(x)}-1}{n^2}}}}$$
where the first identity can be checked by Mathematica and the second is pretty obvious, if you think about undoing the top of the fraction, adding a bit in the middle, then redoing the top.
Next, suppose we have
$$f_k(n)=\frac{\frac{\alpha}2}{n^2+\alpha}+\frac{\frac{\beta}2}{n^2+\beta}$$
where $\alpha=\frac{F_{k}}{F_{k-1}}$ and $\beta=\frac{L_{k}}{L_k-1}$. This is true for $k=2$. We can prove it inductively in a very elegant way by substituting in Binet's formula to get $$\alpha=\frac{\varphi^n-(1-\varphi)^n}{\varphi^{n-1}-(1-\varphi)^{n-1}}$$
$$\beta=\frac{\varphi^n+(1-\varphi)^n}{\varphi^{n-1}+(1-\varphi)^{n-1}}$$
and then plugging the whole mess for $f_k$ into the recurrence relation, setting it equal to $f_{k+1}$ and letting FullSimplify look at it and shrug tell you this is equivalent to a certain rational expression equaling zero, which is, after multiplying out the denominator* $$2 \left(-2+\sqrt{5}\right) \left(\left(6-2 \sqrt{5}\right)^{2 k}-\left(-1+\sqrt{5}\right)^{4 k}\right) e^{i k \pi } n+2 \left(-3+\sqrt{5}\right) \left(\left(6-2 \sqrt{5}\right)^{2 k}-\left(-1+\sqrt{5}\right)^{4 k}\right) e^{i k \pi } n^3+\left(-1+\sqrt{5}\right) \left(\left(6-2 \sqrt{5}\right)^{2 k}-\left(-1+\sqrt{5}\right)^{4 k}\right) e^{i k \pi } n^5=0$$
which looks pretty nasty, and even though Mathematica can't seem to handle it itself, we can solve it pretty easily; it comes down to showing that each coefficient vanishes; in particular we could factor out the following expression from each term
$$\left(6-2 \sqrt{5}\right)^{2 k}-\left(-1+\sqrt{5}\right)^{4 k}$$
however, notice that $(-1+\sqrt{5})^2=6-2\sqrt{5}$ so the expression equals $$(6-2\sqrt{5})^{2k}-(6-2\sqrt{5})^{2k}=0.$$
This, along with the assumption that Mathematica's reductions suffice to complete the inductive proof that $$f_k(n)=\frac{\frac{\alpha}2}{n^2+\alpha}+\frac{\frac{\beta}2}{n^2+\beta}$$
with $\alpha=\frac{F_k}{F_{k-1}}$ and $\beta=\frac{L_k}{L_{k-1}}$.
Then, your conjecture follows immediately from the identity
$$\sum_{n=1}^{\infty}\frac{1}{n^2+c}=\frac{\pi\sqrt{c}\coth(\pi\sqrt{c})-1}{2c}.$$
*The denominator is a non-constant polynomial in $n$, which can only be $0$ at finitely many points for a fixed $k$. Thus this proof shows that the desired equality holds for all but finitely many points, and, since two rational functions agreeing at countably many points are clearly equal, this suffices and we can therefore safely ignore the denominator.
