Error terms for continued fractions of $e^{1\mathrm{/}s}$ and $\sqrt{\frac{v}{u}}\tanh \text{\hspace{0.17em}⁣}\left(\frac{1}{\sqrt{uv}}\right)$

Many classical identities arise from nothing more mysterious than looking at the same object in two different ways. A number, a function, or a combinatorial object may admit several natural decompositions, and by disassembling it in one way and reassembling it in another, we often obtain unexpected corollaries. Telescoping sums provide a particularly vivid incarnation of this principle: by arranging terms so that successive contributions cancel, one performs a conceptual cut-and-paste'' that often admits a clean geometric interpretation. Generating functions offer a complementary perspective. Encoding a problem into a formal power series and then evaluating that series at a prescribed point naturally expresses the same quantity as an infinite (or finite) expansion, and equating these representations yields a wealth of identities. For example, for a real number \(α\) given by its continued fraction expansion $α= [a_0, a_1,a_2,\dots]$, with convergents \(p_n/q_n\) and error terms $E_n := p_n - αq_n$, one can obtain additive’’ decompositions of the form ${\sum }_{n\ge -1}{a}_{n+1}\mid E_n\mid =\alpha +1$, ${\sum }_{n\ge -1}{a}_{n+1}{E}_n^2=\alpha$. Thus $\alpha$ and $\alpha +1$ themselves appear as weighted sums of the local approximation errors of their convergents. In this note we explore what such decompositions yield in two explicit cases: the continued fraction $$e^{1\mathrm{/}s}=[1;\overline{(2k-1)s-1,1,1}{]}_{k=1}^{\mathrm{\infty }}$$ and the continued fraction $$\frac{s}{u}\tanh \text{\hspace{0.17em}⁣}\left(\frac{1}{s}\right)=[0;\overline{(4k-3)u,(4k-1)\frac{s^2}{u}}{]}_{k=1}^{\mathrm{\infty }}$$