Generalized continued fractions can be either represented by Euler's continued fraction formula or Gauss's hypergeometric continued fraction which are both very general.
I will give the OP one simple method which I often use (though I did not adopt it from any textbook but developed it independently)
The idea is that every alternating series can be transformed into a continued fraction.
Given $\sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{a_{n}}$
Let $I_{n}+I_{n+1}=\frac{1}{a_{n}}$ such that
$I_{n+1}+I_{n+2}=\frac{1}{a_{n+1}}$
Then $\frac{I_{n}+I_{n+1}}{I_{n+1}+I_{n+2}}=\frac{a_{n+1}}{a_{n}}$
Now let $r_{n}=\frac{I_{n+1}}{I_{n}}$ such that $r_{n}r_{n+1}=\frac{I_{n+2}}{I_{n}}$
And therefore we have $\frac{1+r_{n}}{r_{n}+r_{n}r_{n+1}}=\frac{a_{n+1}}{a_{n}}$
From which we can deduce the following recurrence relation $$r_{n}=\frac{a_{n}}{a_{n+1}-a_{n}+a_{n+1}r_{n+1}}$$
$$r_{1}=\frac{a_{1}}{a_{2}-a_{1}+a_{2}r_{2}}$$
$$r_{2}=\frac{a_{2}}{a_{3}-a_{2}+a_{3}r_{3}}$$
$$r_{3}=\frac{a_{3}}{a_{4}-a_{3}+a_{4}r_{4}}$$
ad infinitum
After substituting into $\sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{a_{n}}=\frac{1}{a_{1}+a_{1}r_{1}}$
We obtain the following continued fraction formula $$\begin{aligned}\sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{a_{n}}=\cfrac{1}{a_{1}+\cfrac{(a_{1})^2} {a_{2}-a_{1}+\cfrac{(a_{2})^2}{a_{3}-a_{2}+\cfrac{(a_{3})^2}{a_{4}-a_{3}+\cfrac{(a_{4})^2}{a_{5}-a_{4}+\ddots}}}}}\end{aligned}$$
Which can be manipulated and transformed into the following form after equivalence transformation
$$\begin{aligned}\frac{1}{a_{1}}-\frac{1}{a_{1}}\frac{a_{1}}{a_{2}}+
\frac{1}{a_{1}}\frac{a_{1}}{a_{2}}\frac{a_{2}}{a_{3}}+\ddots=\cfrac{\frac{1}{a_{1}}}{1+\cfrac{\frac{a_{1}}{a_{2}}} {1-\frac{a_{1}}{a_{2}}+\cfrac{\frac{a_{2}}{a_{3}}}{1-\frac{a_{2}}{a_{3}}+\cfrac{\frac{a_{3}}{a_{4}}}{1-\frac{a_{3}}{a_{4}}+\ddots}}}}\end{aligned}$$
And finally if we let $$u_{0}=\frac{1}{a_{1}},u_{1}=\frac{a_{1}}{a_{2}},u_{2}=\frac{a_{2}}{a_{3}},u_{3}=\frac{a_{3}}{a_{4}},\ddots$$
We obtain immediately euler's continued fraction formula
$$\begin{aligned}u_{0}-u_{0}u_{1}+u_{0}u_{1}u_{2}-\ddots=\cfrac{u_{0}}{1+\cfrac{u_{1}}{1-u_{1}+\cfrac{u_{2}}{1-u_{2}+\cfrac{u_{3}}{1-u_{3}+\ddots}}}}\end{aligned}$$
P.S :note that in your question you tried to use euler's continued fraction formula for a simple continued fraction(which does not work in general),it only works for generalized continued fractions. In order to determine the convergents of $\phi$ you need to use the fundamental recurrence formulas.
