I discovered a nice Continued Fraction for Higher order roots. (Obviously, there are other and better ways to approximate a higher order root.)
The general theorem is: $\cfrac{\sqrt[n]{A}+1}{\sqrt[n]{A}-1}=1\cdot C+\cfrac{1-(1\cdot n)^2}{3\cdot C +\cfrac{1-(2\cdot n)^2}{5\cdot C+\cfrac{1-(3\cdot n)^2}{7\cdot C+\cfrac{1-(4\cdot n)^2}{9\cdot C+\cfrac{1-(5\cdot n)^2}{11\cdot C+\cfrac{1-(6\cdot n)^2}{13\cdot C+\ddots}}}}}}$
With $C=n\cdot \cfrac{A+1}{A-1}$
Just one example: $\cfrac{\sqrt[5]{11}+1}{\sqrt[5]{11}-1}=1\cdot 6+\cfrac{1-5^2}{3\cdot 6+\cfrac{1-10^2}{5\cdot 6+\cfrac{1-15^2}{7\cdot 6+\cfrac{1-20^2}{9\cdot 6+\cfrac{1-25^2}{11\cdot 6+\ddots}}}}}$
This theorem is a modification of the M. Sardina theorem.
I was able to prove it for n=2 and n=3, applying Euler’s Differential Method but I can not find a general prove.