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.
