Continued fraction for $\frac{1}{e-2}$

A couple of years ago I found the following continued fraction for $\frac1{e-2}$:

$$\frac{1}{e-2} = 1+\cfrac1{2 + \cfrac2{3 + \cfrac3{4 + \cfrac4{5 + \cfrac5{6 + \cfrac6{7 + \cfrac7{\cdots}}}}}}}$$

from fooling around with the well-known continued fraction for $\phi$. Can anyone here help me figure out why this equality holds?

Euler proved in "De Transformatione Serium in Fractiones Continuas" Reference: The Euler Archive, Index number E593 (On the Transformation of Infinite Series to Continued Fractions) [Theorem VI, §40 to §42] that

$$s=\cfrac{1}{1+\cfrac{2}{2+\cfrac{3}{3+\cdots }}}=\dfrac{1}{e-1}.$$

Here is an explanation of how he proceeded.

New Edit in response to @A-Level Student's comment. I transcribed the following assertion from the available translation to English of Euler's article. Now I checked the original paper and corrected equation (1b).

He stated that "we are able to demonstrate without much difficulty, that if

$$\cfrac{a}{a+\cfrac{b}{b+\cfrac{c}{c+\cdots }}}=s,\tag{1a}$$

then

$$a+\cfrac{a}{b+\cfrac{b}{c+\cfrac{c}{d+\cdots }}}=\dfrac{s}{1-s}.\text{"}\tag{1b}$$

Since, in this case, we have $s=1/(1-e)$, $a=1,b=2,c=3,\ldots $ it follows

$$1+\cfrac{1}{2+\cfrac{2}{3+\cfrac{3}{4+\cdots }}}=\dfrac{1}{e-2}.$$

Edit: Euler proves first how to form a continued fraction from an alternating series of a particular type [Theorem VI, §40] and then uses the expansion

$$e^{-1}=1-\dfrac{1}{1}+\dfrac{1}{1\cdot 2}-\dfrac{1}{1\cdot 2\cdot 3}+\ldots .$$

REFERENCES

The Euler Archive, Index number E593, http://www.math.dartmouth.edu/~euler/

Translation of Leonhard Euler's paper by Daniel W. File, The Ohio State University.

Another possibility: remember that the numerators and denominators of successive convergents of a continued fraction can be computed using a three term recurrence.

For a continued fraction

$$b_0+\cfrac{a_1}{b_1+\cfrac{a_2}{b_2+\dots}}$$

with nth convergent $\frac{C_n}{D_n}$, the recurrence

$$\begin{bmatrix}C_n\\\\D_n\end{bmatrix}=b_n\begin{bmatrix}C_{n-1}\\\\D_{n-1}\end{bmatrix}+a_n\begin{bmatrix}C_{n-2}\\\\D_{n-2}\end{bmatrix}$$

with starting values

$\begin{bmatrix}C_{-1}\\\\D_{-1}\end{bmatrix}=\begin{bmatrix}1\\\\0\end{bmatrix}$, $\begin{bmatrix}C_{0}\\\\D_{0}\end{bmatrix}=\begin{bmatrix}b_0\\\\1\end{bmatrix}$

holds.

With $b_j=j+1$ and $a_j=j$, you now try to find a solution for those two difference equations.

Skipping details, the solution of those two recursions are

$$C_n=\frac{(n+3)!}{n+2}\sum_{j=0}^{n+3}\frac{(-1)^j}{j!}$$

and

$$D_n=\frac{(n+3)!}{n+2}\left(1-2\sum_{j=0}^{n+3}\frac{(-1)^j}{j!}\right)$$

are solutions to the two difference equations.

Divide $C_n$ by $D_n$ and take the limit as $n\to\infty$; you should get the expected result.