Is there a general way of finding the value of a continued fraction whose terms form a geometric sequence?

Question

So something like $$\cfrac{1}{2+\cfrac{4}{8+\cfrac{16}{32+\cfrac{64}{\ddots}}}}.$$

Answer 1

(Note: the question title changed, making some of this irrelevant.)

Here is the Wikipedia article on a "continued" fraction (not "continuous").

The article says that $$1+\frac{1}{2+\frac{1}{3+\frac{1}{4+\cdots}}}$$ is equal to $\frac{I_0(2)}{I_1(2)}$ where $I_n(x)$ is a type of Bessel function.

The article gives another example of an arithemetic sequence continued fraction: $$\tanh(1/n)=\frac{1}{n+\frac{1}{3n+\frac{1}{5n+\frac{1}{7n+\cdots}}}}$$

I don't see an example of a geometric sequence continued fraction, and it seems like if something was known about such a thing, even one example, it would be reported here.

Answer 2

Somewhat more generally, note that $$\eqalign{ 1 + \dfrac{r}{r^2 + \dfrac{r^3}{r^4 + r^5/(r^6+\ldots)}} &= 1 + \dfrac{1}{r + \dfrac{r^2}{r^4+r^5/(r^6+\ldots)}}\cr &= 1 + \dfrac{1}{r + \dfrac{1}{r^2 + r^3/(r^6+\ldots)}}\cr &= 1 + \dfrac{1}{r + \dfrac{1}{r^2 + 1/(r^3+\ldots)}}}$$

The convergents of the continued fraction $1+1/(2+1/(4+1/(8+1/(16+\ldots$ are $n(i)/d(i)$ where $n$ and $d$ are OEIS sequences A061377 and A015473 respectively.