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Feb 13, 2021 at 17:00 comment added Pietro Majer Indeed it's the same argument
Feb 13, 2021 at 16:54 history edited GH from MO CC BY-SA 4.0
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Feb 13, 2021 at 16:48 comment added GH from MO Nice elementary argument. It is similar to my answer here: mathoverflow.net/questions/235508/…
Feb 13, 2021 at 15:01 comment added Pietro Majer Exact, $c=0.5931..$
Feb 13, 2021 at 10:34 comment added Mateusz Kwaśnicki In other words: $a_{n+2} \leqslant c a_n$ for $c = \max (\cos^2(x) \cos^2(x+1)) < 1$.
Feb 13, 2021 at 8:47 comment added Pietro Majer (sorry, I had a crush while editing)
Feb 13, 2021 at 8:45 history answered Pietro Majer CC BY-SA 4.0