Deriving a trivial continued fraction for the exponential Lately, I learned about the following continued fraction for the exponential function:
$$\exp(x)=1+\cfrac{x}{1-\cfrac{x/2}{1+x/2-\cfrac{x/3}{1+x/3-\cfrac{x/4}{1+x/4-\dots}}}}$$
I thought it was something new, but evaluating the successive convergents of this continued fraction was a disappointment, as they are nothing more than the partial sums of the usual series $\exp(x)=\sum_{j=0}^{\infty}\frac{x^j}{j!}$.
So, there must be some way to obtain the continued fraction from the series. How might this be done?
 A: This transformation of a series into its equivalent continued fraction,  with the series partial sums being equal to the continued fraction convergents, is due to Euler. The series $$\sum_{n\geq 0}c_{n}=c_0+c_1+\dots+c_n+\dots$$ is transformed into the continued fraction
$$b_0+\mathbf{K}\left( a_{n}|b_{n}\right) =b_0+\dfrac{a_{1|}}{|b_{1}}+\dfrac{a_{2}|}{%
|b_{2}}+\cdots +\dfrac{a_{n}|}{|b_{n}}+\cdots ,$$ 
whose elements $a_{n}$, $b_{n}$ are expressed in terms of $c_n$ as follows:  $b_0=c_0$, $a_1=c_1$, $b_1=1$ and $$a_{n}=-\dfrac{c_{n}}{c_{n-1}},\qquad b_{n}=1+\dfrac{c_{n}}{c_{n-1}}\qquad n\ge 2.$$
For the power series $e^x=\sum_{n\geq 0}\dfrac{1}{n!}x^{n}$, we have  $c_{n}=\dfrac{1}{n!}x^{n}$,
and get $$a_{n}=-\dfrac{c_{n}}{c_{n-1}}x=-\dfrac{1}{n}x,\qquad b_{n}=1+\dfrac{c_{n}}{c_{n-1}}=1+\dfrac{1}{n}x\qquad n\ge 2.$$
Thus
$$\begin{eqnarray*}
e^x &=&\sum_{n\geq 0}\frac{1}{n!}x^{n}=1+\sum_{n\geq 1}\frac{1}{n!}x^{n} \\
&=&1+\frac{x|}{|1}-\frac{\frac{1}{2}x|}{|1+\frac{1}{2}x}-\cdots -\frac{\frac{%
1}{n}x|}{|1+\frac{1}{n}x}-\cdots, 
\end{eqnarray*}$$
which is equivalent to 
$$1+\frac{x|}{|1}-\frac{x|}{|2+x}-\frac{2x|}{|3+x}-\cdots -\frac{nx|}{|n+1+x}+\cdots.$$
This is explained in p.17 of Die Lehre von den Kettenbrüchen Band II by Oskar Perron and proved in Theorem 4.2 of Orthogonal Polynomials and Continued Fractions From Euler´s Point of View by Sergey Khrushchev. It is derived from a theorem that establishes the equivalence between a sequence and a continued fraction.
