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June 9th, 2017, 07:31 AM   #1
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Convergent series : cos(k^2)/k

Hi,

$\displaystyle \textbf{Convergent series? : } \text{the series } \sum \limits_{k=0}^n \frac{\cos(k^2)}{k} \text{ is it convergent?}$

Cordially.

Last edited by skipjack; June 9th, 2017 at 09:24 AM.
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June 9th, 2017, 07:45 AM   #2
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Apparently so (0.444336 ish), but I have no idea how one would prove it.

https://www.wolframalpha.com/input/?...D1+to+infinity
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June 9th, 2017, 08:16 AM   #3
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There is a tip.
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June 9th, 2017, 09:26 AM   #4
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If cosk^2 > a for infinitely many k the series diverges because sum 1/k diverges.

Last edited by skipjack; June 9th, 2017 at 09:01 PM.
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June 9th, 2017, 09:51 AM   #5
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I don't think.

Moreover by using the same reasoning as you, we would have that the sum of cos (k) / k also diverges.
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June 9th, 2017, 12:55 PM   #6
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These series have + and - terms, so it is hard to prove or disprove convergence.
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June 9th, 2017, 12:59 PM   #7
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Quote:
Originally Posted by Dattier View Post
I don't think.

Moreover by using the same reasoning as you, we would have that the sum of cos (k) / k also diverges.
Or indeed $$\sum_{k=1}^\infty \frac{\cos k\pi}{k}$$ which we know converges.
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June 9th, 2017, 01:02 PM   #8
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Quote:
Originally Posted by Dattier View Post
$\displaystyle \textbf{Convergent series? : } \text{the series } \sum \limits_{k=0}^n \frac{\cos(k^2)}{k} \text{ is it convergent?}$
divergent if it starts at $k=0$ ...
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June 9th, 2017, 02:42 PM   #9
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Quote:
Originally Posted by Dattier View Post
$\displaystyle \textbf{Convergent series? : } \text{the series } \sum \limits_{k=1}^n \frac{\cos(k^2)}{k} \text{ is it convergent?}$
I have just realized that in fact my explanation is not good.
So I keep looking, and I throw the thread as soon as I find one, if someone finds one, before he can put it here.

Bye.
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June 19th, 2017, 06:28 AM   #10
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Hi,

here (in French) : un petit exo sympa pour l'agreg. - Page 2

Cordially.
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