Some observations, but not a solution yet.
Let $t_k=\frac{F_{k+1}}{F_k}$ where $F_k$ is the $k^{th}$-Fibonacci number. The convention here for $F_k$ is that $F_3=2, F_4=3, F_5=5, \dots$. Denote the RHS in the above series by $$\Psi_k(x)=\left(\sum_{n=1}^{\infty}\lfloor nt_k\rfloor x^{n-1}\right)^{-1}.$$ Then, it seems that $$\Psi_k(x)=\frac{(1-x)(1-x^{F_k})}{P_k(x)}$$ for some polynomial $P_k(x)$ with the following rather curious properties:
(1) it has degree $F_k-1$;
(2) its coefficients are either $1$ or $2$ (although not clear which is which);
(3) $P_k(1)=F_k$;
(4) $L(t_k)=$ the smallest real root of $P_k(x)$;
(5) there might be a way to relate $(1-x)P_k(x)$ to other $(1-x)P_{\ell}(x)$ for $\ell<k$, recursively.
Here are a few examples:
\begin{align} \Psi_4(x)&=\frac{(1-x)(1-x^3)}{1+2x+2x^3} \\ \Psi_5(x)&=\frac{(1-x)(1-x^5)}{1+2x+x^2+2x^3+2x^4} \\ \Psi_6(x)&=\frac{(1-x)(1-x^8)}{1+2x+x^2+2x^3+2x^4+x^5+2x^6+2x^7} \\ \Psi_7(x)&=\frac{(1-x)(1-x^{12})}{1+2x+x^2+2x^3+2x^4+x^5+2x^6+x^7+2x^8+2x^9+x^{10}+2x^{11}+2x^{12}}. \end{align}