Bernoulli-Carlitz fractions and \(q\)-zeta operators (Q719135)

Let \([n]=(q^n-1)/(q-1)\) for \(n\geq 1\) and let the formal Frobenius operator \(F_n(f)(q)=f(q^n)\) act on the space of formal series with complex coefficients without a constant term \(q\mathbb{C}[[q]]\). To every sequence of complex numbers \((a_n)_{n\geq 1}\) the author associates the \(q\)-Zeta operator \[ \zeta_q(a,s)=\sum_{n\geq 1}\frac{a_n F_n}{[n]^s} \] which acts on \(q\mathbb{C}[[q]]\). This operator is well-defined for every complex \(s\) and the map \(a\rightarrow \zeta_q(a,s)\) between the algebra of complex sequences under convolution and the algebra of linear operators over \(q\mathbb{C}[[q]]\) is a morphism. In particular we are interested in the \(q\)-Zeta operator \[ \zeta_q(s)=\sum_{n\geq 1}\frac{F_n}{[n]^s}\,. \] Define the Bernoulli-Carlitz fractions \(\beta_n\) in the field \(\mathbb{Q}(q)\) by the following recursion. Take \(\beta_0=1\) and \( q(q\beta+1)^n-\beta_n = 1 \) if \(n=1\) and \(q(q\beta+1)^n-\beta_n =0\) if \(n>1\), where after expanding the power of the binomial expression on the place of \(\beta^i\) we put \(\beta_i\). The author proves that for every \(n\geq 2\) \[ \beta_n=\zeta_q(1-n)\left(q-(n+1)q^2\right)\,. \] This is analogous to the classical result for Bernoulli numbers \(B_n=\zeta(1-n)(-n)\) for \(n\geq 2\). The author connects in a similar way other Bernoulli-type of expressions to the images over simple polynomials in \(q\) of operators which correspond to the Riemann zeta function at negative integers.