The formal hyperbolic tangent series and the even series $x\mathrm{/}\tanh x$

Work in $\mathbb{Q}\mathrm{⟦}x\mathrm{⟧}$ with the formal exponential and logarithm of Formal exponential, logarithm, and binomial powers over a commutative $\mathbb{Q}$-algebra and the identities of Formal $\exp$ and $\log$ are inverse homomorphisms and formal binomial powers obey the expected addition laws. Define the formal hyperbolic tangent and the formal area hyperbolic tangent by $T(x):=\frac{\exp (x)-\exp (-x)}{\exp (x)+\exp (-x)}\in \mathbb{Q}\mathrm{⟦}x\mathrm{⟧},A(z):={\sum }_{j\ge 0}\frac{z^{2j+1}}{2j+1}\in \mathbb{Q}\mathrm{⟦}z\mathrm{⟧}\mathrm{.}$

The denominator $\exp (x)+\exp (-x)$ has constant term $2$, a unit of $\mathbb{Q}$, so the quotient is a well-defined power series by A formal power series is a unit exactly when its constant coefficient is a unit; $T(0)=0$ and the linear coefficient of $T$ is $\frac{1}{2}(1-(-1))\mathrm{/}\frac{1}{2}(1+1)=1$. Substituting $-x$ into the defining quotient replaces the numerator by its negative and fixes the denominator, so $T(-x)=-T(x)$: $T$ is odd. Hence $S(x):=\frac{T(x)}{x}\in \mathbb{Q}\mathrm{⟦}x\mathrm{⟧}$ is even with constant term $1$, and $Q(x):=\frac{x}{T(x)}=\frac{1}{S(x)}\in \mathbb{Q}\mathrm{⟦}{x}^2\mathrm{⟧},Q(x)=1+\frac{x^2}{3}-\frac{x^4}{45}+O(x^6)\mathrm{.}$ The displayed coefficients of $Q$ are the normalisation recorded here; they are verified in the inverse-series lemma following on this page, which is the result named in justified_by.

The series $A$ is summable degreewise (Summable families of formal series are locally finite in every coefficient range), $A(0)=0$, and its linear coefficient is $1$; its coefficients are the evaluation of a family whose $j$-th term has order $2j+1$, so no convergence question arises. The symbol $x\mathrm{/}\tanh x$ used by the sources denotes exactly the element $Q(x)$; no analytic convergence, contour, or branch is involved. The residue calculus of Formal Laurent series $K((x))$, their order, derivative, and residue applies to $T$ because $T=xS$ has order $1$. No choice principle is used.