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Mathematica tells the series below doesn't converge. I think it converges. What would the proper way to write things be as an input?

Sum[((-1)^(n) Log[1 + 2 n])/(1 + 2 n), {n, 0, Infinity}]
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By the alternating series test it's clear the series converges, but it seems the convergence is very slow. –  Chris's sis Apr 25 '13 at 18:54
What version of Mathematica are you using? It doesn't tell me that it does not converge. It simply can't calculate it. v9.0.1 here. –  Szabolcs Apr 25 '13 at 18:57
@Szabolcs I use v8.0 here. –  Chris's sis Apr 25 '13 at 18:58
Yes, I can reproduce that in v8. It seems it was a bug in v9, 'fixed' in a certain way in v9. No answer is better than a wrong answer. –  Szabolcs Apr 25 '13 at 18:59
@belisarius We've been talking to ourself for days now... –  Daniel Lichtblau Apr 25 '13 at 21:38

1 Answer 1

To get an actual answer in this case, you could simply retreat to the numerical evaluation of this sum. It works by simply changing Sum to NSum in your code:

NSum[((-1)^(n) Log[1 + 2 n])/(1 + 2 n), {n, 0, Infinity}]

(* ==> -0.192901 *)
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