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March 9th, 2013, 04:47 AM  #1 
Newbie Joined: Mar 2013 Posts: 6 Thanks: 0  Convergence of a sequence
How to prove n/(2 n+sqrt(n)) converges.

March 9th, 2013, 04:49 AM  #2 
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: Convergence of a sequence
Hint : Use L'hospital to see if the limit exists at infinity.

March 9th, 2013, 04:55 AM  #3 
Newbie Joined: Mar 2013 Posts: 6 Thanks: 0  Re: Convergence of a sequence
How do I prove this using the epsilon delta definition of limits?

March 9th, 2013, 04:58 AM  #4  
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: Convergence of a sequence Quote:
 
March 9th, 2013, 05:00 AM  #5 
Newbie Joined: Mar 2013 Posts: 6 Thanks: 0  Re: Convergence of a sequence
I am sorry. I am new to the forum. Where should it be?

March 9th, 2013, 05:09 AM  #6  
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: Convergence of a sequence Quote:
 
March 9th, 2013, 05:10 AM  #7 
Newbie Joined: Mar 2013 Posts: 6 Thanks: 0  Re: Convergence of a sequence
Thank you.

March 9th, 2013, 09:18 AM  #8  
Math Team Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions  Re: Convergence of a sequence Quote:
so we must prove that : we can easily prove that : Hence the sequence : converges ...  
March 9th, 2013, 09:27 AM  #9  
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Convergence of a sequence Quote:
 
March 9th, 2013, 09:29 AM  #10  
Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus  Re: Convergence of a sequence Quote:
I think it is in the right section, as far as I know, sequences with the ?? definition are not taught in high schools. In Greece for example, students in high school, learn about the definition briefly, meaning the they will never use this definition in the final exams and on the other hand only the definion of limits of functions (not sequences, although they are much alike). Speculating from mathbalarka's posts, such things are taught in his school from the elementary grades. Zaidalyafey gave the answer with a better , but here is another way. We will prove that , for every ?>0, we will find a suitable such that . From the last inequality we get . So we proved that for all ?>0 and , , meaning that . [/color]  

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