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Oct 2, 2016 at 8:13 history closed R W
Alexey Ustinov
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Oct 2, 2016 at 2:46 vote accept T. Amdeberhan
Oct 1, 2016 at 23:58 review Close votes
Oct 2, 2016 at 8:13
Oct 1, 2016 at 23:09 history edited GH from MO
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Oct 1, 2016 at 23:08 comment added reuns I would say that $\frac{1}{k \log k} - \int_k^{k+1} \frac{1}{x \log x}dx = \int_k^{k+1}(\frac{1}{k \log k} - \frac{1}{x \log x})dx = \int_k^{k+1} \int_k^x (\frac{-1}{t \log t})' dt dx$ $ = \int_k^{k+1} \int_k^x \frac{1+\log t}{t^2 \log^2 t} dt dx = \mathcal{O}(k^{-2})$ and I think it is more elementary than any calculus-less approach
Oct 1, 2016 at 22:55 answer added Vladimir Dotsenko timeline score: 5
Oct 1, 2016 at 22:21 answer added Fedor Petrov timeline score: 3
Oct 1, 2016 at 21:59 history asked T. Amdeberhan CC BY-SA 3.0