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.
In this answer, I got this continued fraction
$$
\tan(x)=\cfrac{x}{1-\cfrac{x^2}{3-\cfrac{x^2}{\ddots\lower{6pt}{(2n+1)-\cfrac{x^2}{P_n(x)/P_{n+1}(x)}}}}}
$$
where
$$
P_n(x)=\sum_{k=0}^\infty(-x^2)^k\dfrac{2^n(k+n)!/k!}{(2k+2n+1)!}=\left(\frac1x\frac{\mathrm{d}}{\mathrm{d}x}\right)^n\frac{\sin(x)}{x}
$$
I don't know if my proof is any simpler or harder than the one you are considering.
Best Answer
After looking at my previous hint, I was unable to proceed as easily as I thought. Instead, I have here expanded the division of power series in detail.
Start with $$ \begin{align} \frac{\cos(x)}{\sin(x)/x} &=\frac{\sum\limits_{k=0}^\infty(-x^2)^k\dfrac1{(2k)!}} {\sum\limits_{k=0}^\infty(-x^2)^k\dfrac1{(2k+1)!}}\\[12pt] &=\frac{\sum\limits_{k=0}^\infty(-x^2)^k\dfrac{2k+1}{(2k+1)!}} {\sum\limits_{k=0}^\infty(-x^2)^k\dfrac1{(2k+1)!}}\\[12pt] &=1+\frac{\sum\limits_{k=0}^\infty(-x^2)^k\dfrac{2k}{(2k+1)!}} {\sum\limits_{k=0}^\infty(-x^2)^k\dfrac1{(2k+1)!}}\\[12pt] &=1+\frac{\sum\limits_{k=0}^\infty(-x^2)^{k+1}\dfrac{2k+2}{(2k+3)!}} {\sum\limits_{k=0}^\infty(-x^2)^k\dfrac1{(2k+1)!}}\\[12pt] &=1-x^2\left/\left(\frac{\sum\limits_{k=0}^\infty(-x^2)^k\dfrac1{(2k+1)!}} {\sum\limits_{k=0}^\infty(-x^2)^{k}\dfrac{2k+2}{(2k+3)!}}\right)\right.\\[12pt] \end{align} $$ Then note that the ratio of sums is the case $j=0$ of $$ \begin{align} &\frac{\sum\limits_{k=0}^\infty(-x^2)^{k}\dfrac{(2k+2)(2k+4)\dots(2k+2j)}{(2k+2j+1)!}} {\sum\limits_{k=0}^\infty(-x^2)^{k}\dfrac{(2k+2)(2k+4)\dots(2k+2j+2)}{(2k+2j+3)!}}\\[12pt] &=\frac{\sum\limits_{k=0}^\infty(-x^2)^{k}\dfrac{(2k+2)(2k+4)\dots(2k+2j)(2k+2j+2)(2k+2j+3)}{(2k+2j+3)!}} {\sum\limits_{k=0}^\infty(-x^2)^{k}\dfrac{(2k+2)(2k+4)\dots(2k+2j+2)}{(2k+2j+3)!}}\\[12pt] &=(2j+3)+\frac{\sum\limits_{k=0}^\infty(-x^2)^{k}\dfrac{(2k+2)(2k+4)\dots(2k+2j)(2k+2j+2)(2k)}{(2k+2j+3)!}} {\sum\limits_{k=0}^\infty(-x^2)^{k}\dfrac{(2k+2)(2k+4)\dots(2k+2j+2)}{(2k+2j+3)!}}\\[12pt] &=(2j+3)+\frac{\sum\limits_{k=0}^\infty(-x^2)^{k+1}\dfrac{(2k+4)(2k+6)\dots(2k+2j+2)(2k+2j+4)(2k+2)}{(2k+2j+5)!}} {\sum\limits_{k=0}^\infty(-x^2)^{k}\dfrac{(2k+2)(2k+4)\dots(2k+2j+2)}{(2k+2j+3)!}}\\[12pt] &=(2j+3)-x^2\frac{\sum\limits_{k=0}^\infty(-x^2)^{k}\dfrac{(2k+2)(2k+4)\dots(2k+2j+4)}{(2k+2j+5)!}} {\sum\limits_{k=0}^\infty(-x^2)^{k}\dfrac{(2k+2)(2k+4)\dots(2k+2j+2)}{(2k+2j+3)!}}\\[12pt] &=(2j+3)-x^2\left/\left(\frac{\sum\limits_{k=0}^\infty(-x^2)^{k}\dfrac{(2k+2)(2k+4)\dots(2k+2j+2)}{(2k+2j+3)!}} {\sum\limits_{k=0}^\infty(-x^2)^{k}\dfrac{(2k+2)(2k+4)\dots(2k+2j+4)}{(2k+2j+5)!}}\right)\right.\\[12pt] \end{align} $$ and this justifies the continued fraction. That is, if we define $$ P_n(x)=\sum_{k=0}^\infty(-x^2)^k\dfrac{2^n(k+n)!/k!}{(2k+2n+1)!}=\left(\frac1x\frac{\mathrm{d}}{\mathrm{d}x}\right)^n\frac{\sin(x)}{x} $$ we have shown that $$ \tan(x)=\cfrac{x}{1-\cfrac{x^2}{3-\cfrac{x^2}{\ddots\lower{6pt}{(2n+1)-\cfrac{x^2}{P_n(x)/P_{n+1}(x)}}}}} $$