How to use Euler's continued fraction formula? I am trying to convert some continued fractions to series by using Euler's continued fraction formula (see the link to Wikipedia). But there is something I obviously misunderstood in it.
Take for instance the very simple continued fraction for $\phi$ (the golden number; this is a mere example; I am not asking here anything about this number) and look at the section 2, Euler's formula in modern notation, in the Wikipedia page. I can write the number under the expected form; all other requirements are filled. Thus I should write (same notation $r_i$ as in the Wikipedia):
$$
r_i = -\frac{1\times1}{1} = -1
$$
Now I really can't see how I could get $\phi$ by adding integer numbers.
The first convergents should be: $1$, $2$, $3/2$, $5/3$, etc. but I get instead:
$$
1 + (-1) + (-1\times-1) + etc.
$$
I encounter similar issue with all examples I tried.
 A: 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.
A: You were confused because the Wikipedia page in your link had an incorrect result that comes from conflating two different meanings of "b" in the continued fractions literature.  The page has since been updated and the confusing notation has been removed.
When you posted your question, the article said that
$$ r_i = -\frac{a_{i+1} b_{i-1}}{b_{i+1}} \tag{1} \label{1} $$
In the article, $a_i$ and $b_i$ are the $i^\text{th}$ partial numerator and denominator of the fraction.  This result is incorrect.
However, Eq. \eqref{1} is a correct result if $b_i$ is the denominator of the $i^\text{th}$ convergent (i.e., $B_i$ from this article).
$$ r_i = -\frac{a_{i+1} B_{i-1}}{B_{i+1}} \tag{2} \label{2} $$
This can be seen from applying the determinant formula to the difference between the $(n-1)^\text{th}$ and $n^\text{th}$ convergents.
Unfortunately, calculating the $B_i$ explicitly in closed form is usually impossible in practice, so this series representation is not as useful as it might sound.  But in Euler's continued fraction, the partial numerators and denominators are assumed to have a form that causes $B_i=1$, and the series is extremely useful.
