How did Euler obtain this ratio of infinite series from a continued fraction, the terms of the series not being equal to the convergents? If one feels disinclined to read the contextual preamble, I have made some partial progress in clarifying the question. Skip to the bottom! The motivation behind this is to understand the derivation for the simple continued fractions of $\tanh(1/k)$, $e^{2/k}$: Euler’s work has been the only source I could find which attempts to explain the results.
One may find here a translation of one of Euler's essays on continued fractions. Therein, he claims to prove some celebrated continued fraction formulae (with the proof starting in the final sections of the paper). I am finding his work very hard to follow, not from the complexity of the mathematics but more from his very terse style of writing!
I am talking about section 31. For some $a,n$ he considers:

$$s=a+\frac{1}{(1+n)a+\frac{1}{(1+2n)a+\frac{1}{(1+3n)a+\cdots}}}$$

And builds a series of approximations using his formula in section $10,14$: $$a,\frac{(1+n)a^2+1}{(1+n)a},\frac{(1+n)(1+2n)a^3+(2+2n)a}{(1+n)(1+2n)a^2+1},\text{etc.}$$Which are the first few convergents. Now, in section $32$ he continues:

"If these fractions are continued further the law by which they are formed will be easily observed. From this law it may be concluded that after both the numerator and the denominator are divided by the first term of the denominator, the limiting fraction will be:" $$\frac{a+\frac{1}{1\cdot na}+\frac{1}{1\cdot2\cdot1(1+n)n^2a^3}+\frac{1}{1\cdot2\cdot3\cdot1(1+n)(1+2n)n^3a^5}+\text{etc.}}{1+\frac{1}{1(1+n)na^2}+\frac{1}{1\cdot2(1+n)(1+2n)n^2a^4}+\frac{1}{1\cdot2\cdot3(1+n)(1+2n)(1+3n)n^3a^6}+\text{etc.}}$$

Now, he's not wrong - the numerics check out to an excellent accuracy. However I am really confused about where this fraction comes from; no multiplication of terms in the numerator or denominator comes to mind that reconciles that with the sequence of convergents. Moreover, the ratio of truncated series does not exactly equal any of the convergents, so not only do I need to guess the manipulation, but the manipulation leaves you with some kind of vanishing error term. In particular, what he intended the "first term" of the denominator to be, I really don't know - that seems the crux of the matter.
I calculated the above ratio, truncated at the $\cdots$, and compared it to the fourth convergent: $$\frac{\left(1+3n\right)a+3\left(1+2n\right)\left(1+3n\right)na^{3}+6\left(1+n\right)\left(1+2n\right)\left(1+3n\right)n^{2}a^{5}+6\left(1+n\right)\left(1+2n\right)\left(1+3n\right)n^{3}a^{7}}{6\left(1+n\right)\left(1+2n\right)\left(1+3n\right)n^{3}a^{6}+6\left(1+2n\right)\left(1+3n\right)n^{2}a^{4}+3\left(1+3n\right)na^{2}+1}$$Versus the convergent: $$\frac{(1+n)(1+2n)(1+3n)a^4+3(1+n)(1+2n)a^2+1}{(1+n)(1+2n)(1+3n)a^3+2(1+2n)a}$$But I see no clear link.
I would really appreciate it if anyone could comment suggestions or full answers as to what link Euler intended, to obtain the ratio of infinite series.
EDIT: some progress: I discovered through a pattern recognition that the numerators of the convergents $(p_k)_{k=1}^\infty$, indexing from $p_1:=a$, satisfy: $$p_k=\sum_{m=0}^{\lfloor k/2\rfloor}\binom{k-m}{m}\left(\prod_{\nu=m}^{k-m-1}(1+n\nu)\right)a^{k-2m}$$I also find that the denominators $q_k$, indexing with $q_1=1$, satisfy ($k>1$): $$q_k=\sum_{m=0}^{\lfloor(k-1)/2\rfloor}\binom{k-m-1}{m}\left(\prod_{\nu=m+1}^{k-m-1}(1+n\nu)\right)a^{k-2m-1}$$But I haven’t proved it yet, this is just by a comparison.
Hopefully this will help determine the generality of Euler’s expressions (he leaves much hidden behind “$\text{etc.}$”!).
The challenge made concrete: prove that, for any positive integer $a,n$: $$\lim_{k\to\infty}\frac{\sum_{m=0}^{\lfloor k/2\rfloor}\binom{k-m}{m}\left(\prod_{\nu=m}^{k-m-1}\right)a^{k-2m}}{\sum_{m=0}^{\lfloor(k-1)/2\rfloor}\binom{k-m-1}{m}\left(\prod_{\nu=m+1}^{k-m-1}\right)a^{k-2m-1}}=\lim_{N\to\infty}\frac{a+\sum_{m=1}^N\frac{a^{1-2m}}{n^mm!\prod_{\nu=0}^{m-1}(1+n\nu)}}{\sum_{m=0}^N\frac{a^{-2m}}{n^mm!\prod_{\nu=0}^m(1+n\nu)}}$$Which can, in some mysterious way, be made obvious, according to Euler. The curious thing - the two expressions are not equal at any partial $k$ or $N$. As mentioned in the preamble, Euler must have employed something analogous to “creative telescoping”, as some on this site would say, to get a simpler expression modulo a vanishing error term.
 A: It comes as no surprise that Euler's suggestion was the correct one. I found it quite impossible to spot the pattern until everything had been reduced to symbolic form - shame. Anyway, I believe I have figured it out!
So, let $a,n\in\Bbb N$. It is known that: $$\large s=a+\frac{1}{(1+n)a+\frac{1}{(1+2n)a+\frac{1}{(1+3n)a+\frac{1}{(1+4n)a+\cdots}}}}$$Is then guaranteed to converge as the limit of the "convergents" $\frac{p_k}{q_k}$ whose terms are given by the recurrences: $p_1=a,\,p_{k+1}=(1+kn)p_k+p_{k-1}$ and: $q_1=1,\,q_{k+1}=(1+kn)q_k+q_{k-1}$. It was only after manually computing the first $11$ of these that I realised the pattern, with the diagonal entries of "Pascal's" triangle appearing. In general, for $k\gt1$: $$\begin{align}p_k&=\sum_{m=0}^{\lfloor k/2\rfloor}\binom{k-m}{m}\left(\prod_{\nu=m}^{k-m-1}(1+n\nu)\right)a^{k-2m}\\q_k&=\sum_{m=0}^{\lfloor(k-1)/2\rfloor}\binom{k-m-1}{m}\left(\prod_{\nu-m+1}^{k-m-1}(1+n\nu)\right)a^{k-2m-1}\end{align}$$The ratio of which, Euler claims, is asymptotically: $$\large\frac{a+\sum_{m=1}^\infty\frac{a^{1-2m}}{n^m\cdot m!\cdot\prod_{\nu=0}^{m-1}(1+n\nu)}}{\sum_{m=0}^\infty\frac{a^{-2m}}{n^m\cdot m!\cdot\prod_{\nu=0}^m(1+n\nu)}}$$Since $p_k/q_k$ is convergent, it is sufficient to check the result along $k=2N+1$ for $N\in\Bbb N$. The division Euler wanted us to perform was by the $m=0$ term in the $q_k$ sum, $a^{2N}\prod_{\nu=1}^{2N}(1+n\nu)$. Then: $$\large\begin{align}\frac{p_{2N+1}}{q_{2N+1}}&=\frac{a+\sum_{m=1}^Na^{1-2m}\frac{\binom{2N-m+1}{m}}{\prod_{\nu=0}^{m-1}(1+n\nu)\prod_{\nu=2N-m+1}^{2N}(1+n\nu)}}{\sum_{m=0}^Na^{-2m}\frac{\binom{2N-m}{m}}{\prod_{\nu=0}^m(1+n\nu)\prod_{\nu=2N-m+1}^{2N}(1+n\nu)}}\\&=\frac{a+\sum_{m=1}^N\frac{a^{1-2m}}{m!\cdot\prod_{\nu=0}^{m-1}(1+n\nu)}\cdot\color{red}{\frac{(2N-m+1)(2N-m)\cdots(2N-2m+2)}{\prod_{\nu=2N-m+1}^{2N}(1+n\nu)}}}{\sum_{m=0}^N\frac{a^{-2m}}{m!\cdot\prod_{\nu=0}^m(1+n\nu)}\cdot\color{red}{\frac{(2N-m)(2N-m-1)\cdots(2N-2m+1)}{\prod_{\nu=2N-m+1}^{2N}(1+n\nu)}}}\end{align}$$I believe it is rigorous to use the dominated convergence theorem over the counting measure space to conclude the desired result if I show the limit of the two red products, as $N\to\infty$, is $\frac{1}{n^m}$. The two red products are clearly asymptotically similar, so I'll just look at the one on the denominator. I may rewrite it as: $$\frac{\prod_{\nu=0}^{m-1}(2N-m-\nu)}{\prod_{\nu=0}^{m-1}(1+n(2N-\nu))}=\prod_{\nu=0}^{m-1}\frac{1-\frac{m}{2N-\nu}}{n+\frac{1}{2N-\nu}}\overset{N\to\infty}{\longrightarrow}\frac{1}{n^m}$$
And we are done!
