Very interesting question. Right off the bat, it's not even clear to me that the function f(s) defined by a convergent Dirichlet series (when one converges, it tends to be on some right half-plane) necessarily determines the coefficents a(n) uniquely! Assuming convergence at s = 0, it's easy to check that the Kth derivative of f(s) at s = 0 is sum(n=1 to oo) a(n) log(n)^K. (Call this sum b(K).) I'm out of my element with infinite-dimensional operators, but one could say that the (infinite x infinite) matrix M whose (K,L)th entry is given by M(K,L) = log(L)^K, when applied to the infinite vector A = (a(1), a(2), ... ), yields the known infinite vector B = (b(1), b(2), ... ). So I'd say, naively, that if this known matrix M (which is an infinite Vandermonde matrix) has a left inverse M^(-1), then applying M^(-1) to both sides of M A = B should yield A = M^(-1) B, thus solving for all the a(n)'s. --Dan ----------------------------------------------------------------------- Mike Stay asks:

For a standard or exponential generating function, you find the sequence by taking the Taylor series. How does one find the sequence a(n) for a Dirichlet generating function f(s)=sum(n=1 to inf) a(n)/n^s ?