How can I find $\lim \limits_{n\to \infty} \frac{\ln(q_{n+1})}{\ln(q_n)}$? The irrationality measure of a positive real number $x$ can be calculated, if the following limit can be calculated, where $q_n$ are the convergents of the simple continued fraction of $x$.
$$\lim \limits_{n\to \infty} \frac{\ln(q_{n+1})}{\ln(q_n)}$$
Suppose, I have a formula for the entries of the simple continued fraction. Lets say, $a_n=2^n$, so the continued fraction would be $[2,4,8,16,32,\cdots]$.

How can I calculate the above limit ? And how can I find out whether it exists ? Is it possible that the values oscillate and that neither the limit exists nor the sequence diverges to $\infty$ ?

I know the reccurence relation $$q_1=1\ , \,q_2=a_2\ ,\ q_n=a_nq_{n-1}+q_{n-2}\ for\ n>2$$ 
but in general, it will be difficult to find a closed form for $q_n$.
The goal is to find a method to prove special numbers with known continued fraction to be transcendental, for which is enough to show that the above limit is greater than $1$.
 A: Assume $f(n)\leq q_n\leq g(n)$ for some functions $f(n)$ and $g(n)$ such that
$$\lim_{n\to\infty} \frac{g(n)}{f(n)} = C$$
for some real $C$. Then, 
$$\frac{\ln(a_{n+1})}{\ln(g(n))} \leq  \frac{\ln(a_{n+1})}{\ln(q_n)} \leq \frac{\ln(a(n+1))}{\ln(f(n))}$$
Since
$$\lim_{n\to\infty} \frac{g(n)}{f(n)} = C,$$
we have that 
$$\lim_{n\to\infty} \frac{\ln(g(n))}{\ln(f(n))} = \lim_{n\to\infty} \frac{\ln(g(n))}{\ln(g(n))+\ln\left(\frac{f(n)}{g(n)}\right)} = \lim_{n\to\infty} \frac{\ln(g(n))}{\ln(g(n))+\ln(C)}.$$
Because $q_n$ is unbounded, $g(n)$ must also be unbounded, so this limit equals $1$. Thus
$$\lim_{n\to\infty} \frac{\ln(a_{n+1})}{\ln(g(n))} = \lim_{n\to\infty} \frac{\ln(a(n+1))}{\ln(f(n))}$$
which must also be equal to $\lim_{n\to\infty} \frac{\ln(a_{n+1})}{\ln(q_n)}$ by the squeeze theorem. So, we get that 
$$\mu = 2+\lim_{n\to\infty} \frac{\ln(a_{n+1})}{\ln(q_n)} = 2+\lim_{n\to\infty} \frac{\ln(a(n+1))}{\ln(f(n))}.$$
From your definition of $q_n$, we have that, for $n>2$,
$$a_2a_3\cdots a_n \leq q_n \leq a_2(a_3+1)(a_4+1)\cdots (a_n+1).$$
Taking these as $f$ and $g$, we get that
$$\lim_{n\to\infty} \frac{g(n)}{f(n)} = \lim_{n\to\infty} \prod_{k=3}^{n} \frac{a_k+1}{a_k} = \prod_{k=3}^{\infty} \left(1+\frac{1}{a_k}\right).$$
If $a_k$ grows sufficiently, this product will converge, and we can apply our previous result (if not, we must find better bounds $f$ and $g$ - the verification that this product converges is left as an exercise). This gives that
$$\mu = 2+\lim_{n\to\infty} \frac{\ln(a_{n+1})}{\ln(a_2)+\ln(a_3)+\cdots+\ln(a_n)}.$$
Taking $a_n=2^{2^n}$ gives
$$\mu = 2+\lim_{n\to\infty} \frac{2^{n+1}\ln(2)}{\sum_{k=2}^n \ln(2)2^{k}}$$
$$\mu = 2+\lim_{n\to\infty} \frac{2^{n+1}}{2^{n+1}-8}$$
$$\mu = 3.$$
$\mu$ for other sequences can be derived similarly. 
