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Mar 18, 2020 at 23:09 answer added dtg67 timeline score: 7
Dec 5, 2019 at 7:00 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Apr 20, 2019 at 9:38 answer added Alecos Papadopoulos timeline score: 3
Apr 20, 2019 at 9:34 comment added Nick Cox You have the sum of $(1/k) \log k$, each repeated $k$ times. Try this one. What is $1/k$ repeated $k$ times? Just $k (1/k) = 1$.
Apr 20, 2019 at 9:30 review First posts
Apr 20, 2019 at 12:22
Apr 20, 2019 at 9:25 history asked Harman CC BY-SA 4.0