Deriving a trivial continued fraction for the exponential

Solution 1:

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.