Error terms for continued fractions of $e^{1/s}$ and $\sqrt{\frac{v}{u}}\tanh\!\Bigl(\frac{1}{\sqrt{uv}}\Bigr)$

Abstract

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 \(\alpha\) given by its continued fraction expansion $\alpha = [a_0, a_1,a_2,\dots]$, with convergents \(p_n/q_n\) and error terms $E_n := p_n - \alpha q_n$, one can obtain ``additive'' decompositions of the form $\sum_{n\ge-1} a_{n+1}\,\lvert E_n\rvert \;=\; \alpha + 1$, $\sum_{n\ge-1} a_{n+1}\,E_n^{2} \;=\; \alpha$. 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 \[ e^{1/s} = [1;\,{\overline{(2k-1)s-1,1,1}}]_{k=1}^{\infty} \] and the continued fraction \[ \frac{s}{u}\tanh\!\Bigl(\frac{1}{s}\Bigr) = [\,0;\,\overline{(4k-3)u,\,(4k-1)\tfrac{s^{2}}{u}}\,]_{k=1}^{\infty} \]