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.