8 Jun
2007
8 Jun
'07
2:32 p.m.
Hello! Let psi denote the digamma function and gamma=-psi(1). Everyone knows that sum_{n=1}^infnty 1/(n*(1+n*m)) = psi(1 + 1/m) + gamma and that psi(1 + x) + gamma = sum_{k=2}^infnty (-1)^k * zeta(k) * x^(k-1) for x close to zero. What can be said about sum_{n=1}^infnty log(n)/(n*(1+n*m)) and what is the corresponding Maclaurin series? Thank you, Steve Finch http://algo.inria.fr/bsolve/
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