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Nov 16, 2020 at 22:12 comment added ursuv2 Thank you everyone for help. I'm training for some exams and I found it. I lost like 3 hours trying to solve it. Now I got it, thank you
Nov 16, 2020 at 16:31 vote accept ursuv2
Nov 16, 2020 at 13:26 history closed David Roberts
Desiderius Severus
abx
Emil Jeřábek
Todd Trimble
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S Nov 16, 2020 at 9:03 history suggested gmvh
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Nov 16, 2020 at 8:35 review Suggested edits
S Nov 16, 2020 at 9:03
Nov 16, 2020 at 8:02 history edited YCor CC BY-SA 4.0
removed capitals from title
Nov 16, 2020 at 7:52 comment added Pietro Majer Well, indeed the sum telescopizes nicely, but the joy is immediately killed by the suspicion of homeworkness
Nov 16, 2020 at 7:51 history edited Martin Sleziak CC BY-SA 4.0
more descriptive title
Nov 16, 2020 at 7:48 answer added Pietro Majer timeline score: 10
Nov 16, 2020 at 7:33 comment added Dror Speiser Sounds like one of those "it's either trivial, or depends on the irrationality measure of pi being 2" questions :D
Nov 16, 2020 at 4:15 comment added user44191 I'm pretty sure using basic trig and quadratic identities allows us to reduce @ToddTrimble 's method to whether $\sum_{n = 1}^M \sin((n + a)^2)$ and $\sum_{n = 1}^M \cos((n + a)^2)$ are bounded. That said, I suspect that while the question may not be immediately obvious, it seems unlikely to be asked with research-level understanding.
Nov 16, 2020 at 1:34 comment added Todd Trimble Convergence would follow if sums of the form $\sum_{n=1}^M \sin(n)\sin(n^2)$ had an upper bound independent of $M$. This is certainly known for sums $\sum_{n=1}^M \sin(n)$. I'm sure it's also true for these sums, but I don't have an immediate proof.
Nov 16, 2020 at 1:05 review Close votes
Nov 16, 2020 at 13:31
Nov 16, 2020 at 1:02 history edited David Roberts CC BY-SA 4.0
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Nov 16, 2020 at 0:49 history edited ursuv2 CC BY-SA 4.0
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Nov 16, 2020 at 0:41 history edited Konstantinos Kanakoglou CC BY-SA 4.0
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Nov 16, 2020 at 0:31 review First posts
Nov 16, 2020 at 2:16
Nov 16, 2020 at 0:29 history asked ursuv2 CC BY-SA 4.0