A new series for π3 and related congruences

Let H-n((2)) denote the second-order harmonic number Sigma(0<k <= n) 1/k(2) for n = 0, 1, 2,.... In this paper we obtain the following identity: Sigma(infinity)(k=1) 2(k)H(k-1)((2))/k((2k)(k)) = pi(3)/48. We explain how we found the series and develop related congruences involving Bernoulli or Euler numbers; for example, it is shown that Sigma(p-1)(k=1) ((2k)(k))/2(k)H(k)((2)) equivalent to -Ep-3 (mod p) for any prime p > 3, where E-0, E-1, E-2,... are Euler numbers. Motivated by the Amdeberhan-Zeilberger identity Sigma(infinity)(k=1) (21k - 8)/(k(3)((2k)(k))(3)) = pi(2)/6, we also establish the congruence Sigma((p-1)/2)(k=1) 21k-8/k(3)((2k)(k))(3) equivalent to (-1)((p+1)/2)4E(p-3) (mod p) for each prime p > 3.