(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...) |
|||
| (3 intermediate revisions by 3 users not shown) | |||
| Line 1: | Line 1: | ||
| − | <math>\ | + | <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 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]] | ||
| + | |||
| + | 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]] | ||
| + | |||
| + | 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 | ||
| + | <math>x \tan \frac{1}{x}</math> | ||
| + | ...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 21:31, 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)
