Closed form solution for ∑∞n=111+n21+1⋱1+11+n2\sum_{n=1}^\infty\frac{1}{1+\frac{n^2}{1+\frac{1}{\stackrel{\ddots}{1+\frac{1}{1+n^2}}}}}.

This is again only a part of a proof, because I believe I made a mistake somewhere along the line. Anyway, we’ll start from this equation
$$\frac{1}{1+\frac{n^2}{1+\frac{1}{1+\frac{\ddots}{1+\frac{1}{1+\frac{1}{1+n^2}}}}}}=\frac{F_{k-1}\:n^2+F_{k}}{F_{k-2}\:n^4+2F_{k-1}\:n^2+F_{k}}$$
where k is equal to the number of horizontal lines on the left hand side. user109899 went off of the same premise, but we’ll change it up a little. Let’s start by partially factoring the denominator on the right hand side.
$$\frac{1}{F_{k-2}\:n^4+2F_{k-1}\:n^2+F_{k}}=\frac{1}{(F_{(k-2)/2}\:n^2+F_{k/2})(L_{(k-2)/2}\:n^2+L_{k/2})}$$
This follows from the fact that
$$F_{n}+F_{n+2}=L_{n+1}$$
Now we need the complex roots of each term. On the left hand side, these are
$$\pm \:i\sqrt{\frac{F_{k/2}}{F_{(k-2)/2}}}$$
And on the right side these are just
$$\pm \:i\sqrt{\frac{L_{k/2}}{L_{(k-2)/2}}}$$
At this point I’m also going to define a couple of variables to represent these roots in a way that we can more easily work with.
$$a_1=\sqrt{\frac{F_{k/2}}{F_{(k-2)/2}}}$$
$$a_2=\sqrt{\frac{L_{k/2}}{L_{(k-2)/2}}}$$
If we plug these roots back into the equation from the first step, we get an equation that looks like this
$$f(n)=\frac{F_{k-1}\:n^2+F_{k}}{(n+ia_1)(n-ia_1)(n+ia_2)(n-ia_2)}$$
For a more detailed description of this next step, read [this][1] thread. I won’t explain this much since it’s already shown there, but we need to find the residues of an equation $g(z):=\pi cot(\pi z)\:f(n)$ with poles at $\pm\:ia_1$ and $\pm\:ia_2$ (Note that the residue for each pole is identical to its opposite pole because $\coth$ is an odd function).
$$b_{a1}=\frac{\pi(F_{k-1}\:a_1^2-F_{k})\:\coth(\pi a_1)}{2a_1(ia_1+ia_2)(ia_1-ia_2)}=-\frac{\pi(F_{k-1}\:a_1^2-F_{k})\:\coth(\pi a_1)}{2a_1(a_1^2-a_2^2)}$$
$$b_{a2}=\frac{\pi(F_{k-1}\:a_2^2-F_{k})\:\coth(\pi a_2)}{2a_2(ia_2+ia_1)(ia_2-ia_1)}=-\frac{\pi(F_{k-1}\:a_2^2-F_{k})\:\coth(\pi a_2)}{2a_2(a_2^2-a_1^2)}$$
We can then plug $f(x)$ into an infinite sum to evaluate it based upon these residues.
$$\sum_{n=-\infty}^{\infty}\frac{F_{k-1}\:n^2+F_{k}}{(n^2+a_1^2)(n^2+a_2^2)}=-(2b_{a1}+2b_{a2})=$$ $$\frac{\pi(F_{k-1}\:a_1^2-F_{k})\:\coth(\pi a_1)}{a_1(a_1^2-a_2^2)}+\frac{\pi(F_{k-1}\:a_2^2-F_{k})\:\coth(\pi a_2)}{a_2(a_2^2-a_1^2)}$$
Using the identity $F_k-F_{k+1}\frac{F_{k/2}}{F_{(k+1)/2}}=-1$ and $F_k-F_{k+1}\frac{L_{k/2}}{L_{(k+1)/2}}=1$ for odd k, we can simplify this a bit further.
$$\small{\pi\left (F_{k-1}-\frac{F_{k}}{a_1^2}-\frac{F_{k-1}\:a_1^2-F_{k}}{a_2^2}\right )\:a_1\coth(\pi a_1)+\pi\left (F_{k-1}-\frac{F_{k}}{a_2^2}-\frac{F_{k-1}\:a_2^2-F_{k}}{a_1^2}\right )\:a_2\coth(\pi a_2)=}$$ $$\pi\left (\left (-1-\frac{F_{k-1}\:a_1^2-F_{k}}{a_2^2}\right )a_1\coth(\pi a_1)+\left (1-\frac{F_{k-1}\:a_2^2-F_{k}}{a_1^2}\right )a_2\coth(\pi a_2)\right )$$
Because the original function is symmetric for negative $n$, we can subtract the zero term and divide the result by two in order to get the sum from 1 to infinity.
$$\frac{F_{k-1}\:0^2+F_{k}}{F_{k-2}\:0^4+2F_{k-1}\:0^2+F_{k}} = 1$$
$$\small{\sum_{n=1}^{\infty}\frac{F_{k-1}\:n^2+F_{k}}{(n^2+a_1^2)(n^2+a_2^2)}=\frac{\pi\left (\left (-1-\frac{F_{k-1}\:a_1^2-F_{k}}{a_2^2}\right )a_1\coth(\pi a_1)+\left (1-\frac{F_{k-1}\:a_2^2-F_{k}}{a_1^2}\right )a_2\coth(\pi a_2)\right ) - 1}{2}}=$$
$$\small{\frac{\pi}{2}\left (\left (-1-\frac{F_{k-1}\:a_1^2-F_{k}}{a_2^2}\right )a_1\coth(\pi a_1)+\left (1-\frac{F_{k-1}\:a_2^2-F_{k}}{a_1^2}\right )a_2\coth(\pi a_2)\right ) - \frac{1}{2}}$$
Not-so-simple algebra and applying certain properties of fibonacci numbers shows that
$$-1-\frac{F_{k-1}\:\frac{F_{k/2}}{F_{k/2-1}}-F_{k}}{\frac{L_{k/2}}{L_{k/2-1}}}=-\frac{2}{F_{k-1}+1}$$ and similarly $$1-\frac{F_{k-1}\:\frac{L_{k/2}}{L_{k/2-1}}-F_{k}}{\frac{F_{k/2}}{F_{k/2-1}}}=-\frac{2F_{k-1}}{F_{k-1}-1}$$
Plugging these values into the equation now gives
$$\frac{\pi}{2}\left (\left (-\frac{2}{F_{k-1}+1}\right )a_1\coth(\pi a_1)+\left (-\frac{2F_{k-1}}{F_{k-1}-1}\right )a_2\coth(\pi a_2)\right ) - \frac{1}{2}$$
At this point, if we expand the variables, we get
$$\small{\frac{\pi}{2}\left (\left (-\frac{2}{F_{k-1}+1}\right )\sqrt{\frac{F_{k/2}}{F_{k/2-1}}}\coth\left (\sqrt{\frac{F_{k/2}}{F_{k/2-1}}}\pi \right )+\left (-\frac{2F_{k-1}}{F_{k-1}-1}\right )\sqrt{\frac{L_{k/2}}{L_{k/2-1}}}\coth\left (\sqrt{\frac{L_{k/2}}{L_{k/2-1}}}\pi \right )\right ) - \frac{1}{2}}$$
Which is very similar to your original equation, but is not the same. I think I must have made a mistake in my math somewhere, but hopefully this helps in finding a correct proof. I’ve spent over half a day on this equation now, though, so I’ll leave that to someone else 🙂