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June 9th, 2017, 08:31 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 10:24 AM. 
June 9th, 2017, 08:45 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,118 Thanks: 2369 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, 09:16 AM  #3 
Member Joined: May 2017 From: France Posts: 57 Thanks: 1 
There is a tip.

June 9th, 2017, 10:26 AM  #4 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,214 Thanks: 91 
If cosk^2 > a for infinitely many k the series diverges because sum 1/k diverges.
Last edited by skipjack; June 9th, 2017 at 10:01 PM. 
June 9th, 2017, 10:51 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, 01:55 PM  #6 
Global Moderator Joined: May 2007 Posts: 6,416 Thanks: 558 
These series have + and  terms, so it is hard to prove or disprove convergence.

June 9th, 2017, 01:59 PM  #7 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,118 Thanks: 2369 Math Focus: Mainly analysis and algebra  
June 9th, 2017, 02:02 PM  #8 
Math Team Joined: Jul 2011 From: Texas Posts: 2,692 Thanks: 1351  
June 9th, 2017, 03:42 PM  #9  
Member Joined: May 2017 From: France Posts: 57 Thanks: 1  Quote:
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, 07:28 AM  #10 
Member Joined: May 2017 From: France Posts: 57 Thanks: 1  

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