consider an arbitrary continued fraction, do we have an theorem about its convergence or conditions about converging?
the question is came from examples like $$ x=5-\cfrac{6}{5-\cfrac{6}{5-\cfrac{6}{\ddots}}}$$
which $x$ can be both $2$ and $3$ ! I think there is special conditions that make it to converge to one of these numbers. it will be great if you help me with this. thanks a lot.
On
I don't understand why you think that the fixpoints of $$ f(x) = 5 - \frac{6}{x} $$ are 5 and 1. They are actually 2 and 3. The derivative of the function at 2 has absolute value larger than 1, so unstable. The derivative at 3 has absolute value below 1, stable fixpoint.
You should be able to prove, say for $x > 2,$ that $$ \left| \left( 5 - \frac{6}{x} \right) - 3 \right| < \left| x - 3 \right| $$
More than that, you should be able to find a constant $0 < C < 1$ such that, for $x > \frac{5}{2},$ $$ \left| \left( 5 - \frac{6}{x} \right) - 3 \right| \leq C \left| x - 3 \right| $$ The meaning of the 5/2 is that the derivative is 1 at $\sqrt 6 \approx 2.449. $ The Mean Value Theorem is probably the quickest way to argue this, and works best if $x$ is bounded strictly away from $\sqrt 6.$ So, take $x > 5/2.$
Yep, for $x > 0,$ $f'' < 0$ so for $x > 5/2,$ $f'(x) < f'(5/2) = 24/25.$ So, we may take $$C = \frac{24}{25}$$
As Will already explained, the answer in this case is 3. As a matter of fact if you take almost any number $x$ and define sequence $(a_k)$ by the following rules $$ a_0 = x,\ a_{k+1} = 5 - \frac6{a_k} $$ then it converges to 3. In the case of our continued fraction we have $x=5$ and you can prove (it is quite easy) that $b_k = a_k-3$ is positive and monotonically decreasing, and not just decreasing but doing it pretty fast -- $b_{k+1} < \frac23 b_k$ -- from which it immediately follows that the limit is 3.
Similar methods can be applied to many other generalized continued fractions of the same type $$ a_k = p - \frac{q}{p-\frac{q}{p-\frac{q}{...}}} $$ but perhaps you would like to explore that by yourself. Good luck!