The question seems to be about $[1;2,3,3,3,3,3,\dots]$. A simple approach, assuming that the calculation makes sense, is to let $\alpha=[0;3,3,3,3,3,\dots]$.
If you look at this continued fraction, you can see that $\alpha=\frac{1}{3+\alpha}$. Rewrite as $\alpha^2+3\alpha-1=0$ and solve for $\alpha$. Now it is easy to compute $[1;2,3,3,3,3,3,3,\dots]$. For $[1;2,3,3,3,3,3,3,\dots]=1+\cfrac{1}{2+\alpha}.$
Details: So $\alpha=\frac{-3+\sqrt{13}}{2}$, and our original continued fraction is $\frac{3+\alpha}{2+\alpha}$. After a bit of fiddling, this turns out to be $\frac{3+\sqrt{13}}{1+\sqrt{13}}$. That is a correct answer, but you probably don't want a square root in the denominator, so multiply numerator and denominator by $\sqrt{13}-1$.
Remark: A similar process applies, in principle, to any ultimately cycling continued fraction. In all cases we get a quadratic irrationality.