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Note: There is no simpler way compared to what you have already indicated. But, in fact it's not more complicated than the presumably more familiar Cauchy product of polynomials. Let's make a comparison head-to-head: We consider two Dirichletpolynomials \begin{align*} f_1(s)=\sum_{n=1}^m\frac{a_n}{n^s} \qquad \text{ and } \qquad ...



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