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April 21st, 2016, 07:21 PM  #1 
Member Joined: May 2015 From: Earth Posts: 64 Thanks: 0  Using limit comparison with negative numbers?
So i'm looking at an answer sheet to an old exam, and it's asking for me to prove that the sum of (n)/(2n^2+5) converges or diverges. They use limit comparison in the answer. Everywhere I look says that the parts in the sum must be greater than 0, yet the answer sheet seems to say differently? It gives ( (n)/(2n^2+5) )/( 1/n ) as n > infinity as the setup to the problem. Does the fact that this fraction is not negative make this okay? These series are killing me and nothing really makes sense. 
April 21st, 2016, 07:56 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,448 Thanks: 2499 Math Focus: Mainly analysis and algebra 
$$\sum_{n=1}^\infty {n \over 2n^2+5} = \sum_{n=1}^\infty {n \over 2n^2 + 5}$$ So the sum on the left hand side converges if and only if the sum on the right hand side converges. 
April 21st, 2016, 08:14 PM  #3 
Member Joined: May 2015 From: Earth Posts: 64 Thanks: 0 
That's helpful, but that's not what was written in the answer key. I understand why it was okay for them to change use the negative signs in their work, since they cancel out, but is that the only reason it's acceptable? Still seems weird.

April 21st, 2016, 08:26 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,448 Thanks: 2499 Math Focus: Mainly analysis and algebra 
Yes, they are using $\frac1x$ because the negatives cancel out to give a finite limit greater than zero which is the definition of the test. Personally, I would say it's more correct to do the transformation above on both series to make the terms positive, and then run the test. Last edited by v8archie; April 21st, 2016 at 08:44 PM. 
April 22nd, 2016, 10:38 AM  #5 
Member Joined: May 2015 From: Earth Posts: 64 Thanks: 0 
Okay that makes a lot more sense, thank you!!


Tags 
comparison, limit, negative, numbers 
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