The Catalan generating function $C(x)={\sum }_{n\ge 0}{C}_n{x}^n$ in $\mathbb{Q}\mathrm{⟦}x\mathrm{⟧}$

$\mathbb{Q}$ is a field (The rationals form a field) and therefore a commutative ring (Every field is a commutative ring with $1\mathrm{\ne }0$; it is an integral domain, and it is a commutative division ring), so the formal power series $\mathbb{Q}\mathrm{⟦}x\mathrm{⟧}$ and the coefficient functionals $[x^n]$ of Formal power series over a commutative ring and the coefficient-extraction functional $[x^n]$ are available over it.

Natural numbers as coefficients. A natural number written where a rational is expected denotes its image under the composite of the embedding $\mathbb{N}\to \mathbb{Z}$, $k\mapsto [(k,0)]$, of The naturals embed in the integers with the embedding $\mathbb{Z}\to \mathbb{Q}$ of The integers embed in the rationals; no symbol is written for it. Both embeddings are injective and preserve addition and multiplication, so the composite does too, and by induction (The principle of mathematical induction) it therefore carries a finite sum or product of natural numbers to the corresponding finite sum or product of rationals. An identity between natural numbers may therefore be read as an identity between rationals, and conversely, the embedding being injective.

Definition. The Catalan generating function is the formal power series $C\in \mathbb{Q}\mathrm{⟦}x\mathrm{⟧}$ whose coefficient function is $n\mapsto C_n$ (The Catalan number $C_n:=\mid {\mathcal{D}}_n\mid$), that is

$C={\sum }_{n\ge 0}{C}_n{x}^n,[x^n]C=C_n(n\in \mathbb{N})\mathrm{.}$

Two formal power series are equal exactly when all their coefficients agree (Coefficient extraction is $R$-linear, separates formal series, shifts under multiplication by $x^k$, and converts products to finite convolution), so $C$ is determined by this prescription and nothing else is asserted: the symbol $x$ is an indeterminate, no value is substituted for it, and no convergence is claimed.

$\mathbb{Q}\mathrm{⟦}x\mathrm{⟧}$ as a commutative $\mathbb{Q}$-algebra. The coefficientwise sum and the Cauchy product make $\mathbb{Q}\mathrm{⟦}x\mathrm{⟧}$ a commutative ring, and the map sending a rational to the constant series with that coefficient at $0$ is an injective unital ring homomorphism $\mathbb{Q}\to \mathbb{Q}\mathrm{⟦}x\mathrm{⟧}$ (Cauchy multiplication makes $R\mathrm{⟦}x\mathrm{⟧}$ a commutative ring containing $R[x]$ as the finitely supported subring, applied to the polynomials of degree at most $0$). So $\mathbb{Q}\mathrm{⟦}x\mathrm{⟧}$ is a commutative $\mathbb{Q}$-algebra in the sense of Formal exponential, logarithm, and binomial powers over a commutative $\mathbb{Q}$-algebra, and the formal exponential, logarithm and binomial powers of that item are available in it.

• Why $\mathbb{Q}$ and not $\mathbb{Z}$. Every coefficient of $C$ is a natural number, so $C$ has a copy in $\mathbb{Z}\mathrm{⟦}x\mathrm{⟧}$. The square-root and binomial-power machinery used below is stated for a commutative $\mathbb{Q}$-algebra, because its definitions divide by $n!$, and $\mathbb{Z}\mathrm{⟦}x\mathrm{⟧}$ is not one. Working over $\mathbb{Q}$ from the start avoids moving between two rings in the middle of a computation.

• A count read as a coefficient. The coefficients are the counts $C_n=\mid {\mathcal{D}}_n\mid$, and the reading of a natural number as a rational is the embedding recorded above. Nothing else changes: an identity proved between the counts is an identity between the coefficients, and an identity proved between the coefficients transports back because the embedding is injective.