I arrived at this while solving for a Physics Problem. $$1+\frac{1}{\frac{1}{2}+\frac{1}{3+\frac{1}{\frac{1}{4}+\frac{1}{5+\frac{1}{\frac{1}{6}+\frac{1}{7+\frac{1}{\frac{1}{8}+\frac{1}{9+\frac{1}{\frac{1}{10}+...}}}}}}}}}$$
I was wondering whether this has a closed form.
I think the notation for this is, $$[1:\frac{1}{2},3,\frac{1}{4},5,\frac{1}{6},...,]$$ I was not able to find much about it online.
EDIT:
This was the Physics Problem if anyone interested, though the physics is superficial here.
This expression can be evaluated with Gauss' continued fraction : \begin{equation} \frac{\mathbf{F}\left(a,b;c;z\right)}{\mathbf{F}\left(a,b+1;c+1;z\right)}=t_{0% }-\cfrac{u_{1}z}{t_{1}-\cfrac{u_{2}z}{t_{2}-\cfrac{u_{3}z}{t_{3}-\cdots}}} \end{equation} where \begin{align} t_n&=c+n\\ u_{2n+1}&=(a+n)(c-b+n)\\ u_{2n}&=(b+n)(c-a+n) \end{align} and $\mathbf{F}$ is the regularized hypergeometric function.
In view of Gauss' formula, one may transform first \begin{equation} L=1+\frac{1}{\frac{1}{2}+\frac{1}{3+\frac{1}{\frac{1}{4}+\frac{1}{5+\frac{1}{\frac{1}{6}+\frac{1}{7+\frac{1}{\frac{1}{8}+\frac{1}{9+\frac{1}{\frac{1}{10}+...}}}}}}}}} \end{equation} into an equivalent expression with linearly increasing coefficients $t_n$: \begin{equation} L=1+\frac{4}{2+\frac{4}{3+\frac{16}{4+\frac{16}{5+\frac{36}{6+\frac{36}{7+\frac{64}{8+\frac{64}{9+\frac{100}{10+...}}}}}}}}} \end{equation} then, by choosing $c=1, a=1, b=0,z=-4$, we have \begin{align} t_n&=n+1\\ u_{2n+1}&=(n+1)^2\\ u_{2n}&=n^2 \end{align} and thus \begin{equation} L=\frac{{}_2F_1(1,0;1,-4)}{{}_2F_1(1,1;2,-4)}=\frac{1}{{}_2F_1(1,1;2,-4)} \end{equation} Since (DLMF) \begin{equation} {}_2F_1\left(1,1;2;z\right)=-z^{-1}\ln\left(1-z\right) \end{equation} it comes \begin{equation} L=\frac{4}{\ln5} \end{equation} as expected from the numerical evaluations in the comments.