The g.f. of these numbers (see the link) is $\sum_{n=0}^\infty S_nx^n=\frac{1-x-\sqrt{1-6x+x^2}}{2x}$. Thus radius of convergence is the same of the radical, that is the modulus of the smaller root of $1-6x+x^2$, which is $3-\sqrt{8}=\lim_{n\to\infty}\frac{S_{n-1}}{S_n}$.
(edit 12/3/21). I feel obliged to improve a little this answer. The above argument shows that if the limit of the ratio $S_{n-1}/S_{n}$ exists, its value is $3-\sqrt{8}$. However, it is not immediately clear that there should be a limit (e.g. the ratio of coefficients of $\frac {1+ax}{1-x^2}$ alternates between $a $ and $1/a$). Here is an elementary existence argument. From the g.f. one gets the two-term recurrence $$S_n=\frac{6n-3}{n+1}S_{n-1}-\frac{n-2}{n+1}S_{n-2}$$ So $\rho_n:=S_n/S_{n-1}$ satisfies $$\cases{\rho_1=2\\\\\rho_n=\frac{6n-3}{n+1}-\frac{n-2}{n+1}\frac1{\rho_{n-1}}.}$$ It follows easily by induction e.g. $1\le \rho_n\le 6$. Since $x\mapsto-\frac1x$ is increasing we have $ \liminf_{n\to\infty}(-1/\rho_{n-1})=-1/{\liminf_{n\to\infty}\rho_{n}}$ and $ \limsup_{n\to\infty}(-1/\rho_{n-1})=-1/{\limsup_{n\to\infty}\rho_{n}}$. Therefore both $\liminf_{n\to\infty}\rho_n$ and $\limsup_{n\to\infty}\rho_n$ solve the fixed point equation $\lambda=6-\frac1\lambda$; lying in the interval $[1,6]$ they both coincide with $3+\sqrt{8}$, which is therefore the limit. (By a little more computation or more thinking one should be able to prove that $\rho_n$ in fact increasing).