(New page: <math>\sum-{n = 1}^\infty n \tan \frac{1}{n}</math> I'm fairly certain that this can't be integrated. I tried finding a sum with terms greater than this one that converged, but haven't h...)
 
 
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I'm fairly certain that this can't be integrated.  I tried finding a sum with terms greater than this one that converged, but haven't had any luck.  Any thoughts or hints? --[[User:Jmason|John Mason]]
 
I'm fairly certain that this can't be integrated.  I tried finding a sum with terms greater than this one that converged, but haven't had any luck.  Any thoughts or hints? --[[User:Jmason|John Mason]]
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Yeah, don't try to integrate it.  Use the nth term test instead.  That will be much more effective.  :)  I tried to go through the whole integrating thing too and it got me nowhere. Then I looked at the terms of the sequence and realized I could use the nth term test. [[User:Jhunsber|His Awesomeness, Josh Hunsberger]]
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Funny story...I immediately recognized this problem because Thursday night, I was helping a friend in 165 with her homework, and she is studying L'Hopital's rule.  She was to find the limit as x -> Infinity of
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<math>x \tan \frac{1}{x}</math>
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...and we couldn't figure out how to do it immediately.  So because this specific problem gave me issues only a couple nights ago, I remembered that the limit equaled 1, and thus not 0, without much thinking at all.  Ha ha. --[[User:Reckman|Randy Eckman]] 01:30, 3 November 2008 (UTC)

Latest revision as of 20:31, 2 November 2008

$ \sum_{n = 1}^\infty n \tan \frac{1}{n} $

I'm fairly certain that this can't be integrated. I tried finding a sum with terms greater than this one that converged, but haven't had any luck. Any thoughts or hints? --John Mason

Yeah, don't try to integrate it. Use the nth term test instead. That will be much more effective.  :) I tried to go through the whole integrating thing too and it got me nowhere. Then I looked at the terms of the sequence and realized I could use the nth term test. His Awesomeness, Josh Hunsberger

Funny story...I immediately recognized this problem because Thursday night, I was helping a friend in 165 with her homework, and she is studying L'Hopital's rule. She was to find the limit as x -> Infinity of $ x \tan \frac{1}{x} $ ...and we couldn't figure out how to do it immediately. So because this specific problem gave me issues only a couple nights ago, I remembered that the limit equaled 1, and thus not 0, without much thinking at all. Ha ha. --Randy Eckman 01:30, 3 November 2008 (UTC)

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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett