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You can sum your infinite series by shifting the sum over n along by 1/2, which then simplifies the summand to a form that Mathematica can handle. (1/(-1)^(1/2)) Sum[((-1)^n Log[2 n])/(2 n), {n, 1/2, Infinity, 1}] (* (1/4) (Pi Log[4] + StieltjesGamma[1, 1/4] - StieltjesGamma[1, 3/4]) *) The overall factor 1/(-1)^(1/2) cancels the (-1)^(1/2) factor that ...



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