My Math Forum Convergent series : cos(k^2)/k

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 June 9th, 2017, 07:31 AM #1 Member   Joined: May 2017 From: France Posts: 57 Thanks: 1 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.
 June 9th, 2017, 07:45 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,309 Thanks: 2444 Math Focus: Mainly analysis and algebra Apparently so (0.444336 ish), but I have no idea how one would prove it. https://www.wolframalpha.com/input/?...D1+to+infinity
 June 9th, 2017, 08:16 AM #3 Member   Joined: May 2017 From: France Posts: 57 Thanks: 1 There is a tip.
 June 9th, 2017, 09:26 AM #4 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 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.
 June 9th, 2017, 09:51 AM #5 Member   Joined: May 2017 From: France Posts: 57 Thanks: 1 I don't think. Moreover by using the same reasoning as you, we would have that the sum of cos (k) / k also diverges.
 June 9th, 2017, 12:55 PM #6 Global Moderator   Joined: May 2007 Posts: 6,512 Thanks: 585 These series have + and - terms, so it is hard to prove or disprove convergence.
June 9th, 2017, 12:59 PM   #7
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Quote:
 Originally Posted by Dattier 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.

June 9th, 2017, 01:02 PM   #8
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Quote:
 Originally Posted by Dattier $\displaystyle \textbf{Convergent series? : } \text{the series } \sum \limits_{k=0}^n \frac{\cos(k^2)}{k} \text{ is it convergent?}$
divergent if it starts at $k=0$ ...

June 9th, 2017, 02:42 PM   #9
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Quote:
 Originally Posted by Dattier $\displaystyle \textbf{Convergent series? : } \text{the series } \sum \limits_{k=1}^n \frac{\cos(k^2)}{k} \text{ is it convergent?}$
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.

 June 19th, 2017, 06:28 AM #10 Member   Joined: May 2017 From: France Posts: 57 Thanks: 1 Hi, here (in French) : un petit exo sympa pour l'agreg. - Page 2 Cordially.

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