Is there a closed-form expression or a very good approximation for $$ \sum_{i=0}^n \binom{n}{i} \log (i+1) \,? $$ If the summands alternate, then there is a very close approximation, yet it feels like the alternation is a crucial ingredient.
So I was wondering if one can do better than estimating the sum by $\log(n+1) 2^n$.

  • $\begingroup$ The question you cite doesn't give a closed form, and (by the discussion) a closed form for the alternating sum is extremely unlikely. $\endgroup$ – vonbrand Mar 14 '14 at 10:07
  • $\begingroup$ @vonbrand OK, I changed the wording. The approximation in the linked question is very precise, that would be OK too. $\endgroup$ – john_leo Mar 14 '14 at 10:23

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