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I got that it diverged because the limit went to infinity...maybe I'll just ask about it in class and see where I went wrong, if I am wrong, which is more than likely. --[[User:Reckman|Randy Eckman]] 12:19, 5 November 2008 (UTC)
 
I got that it diverged because the limit went to infinity...maybe I'll just ask about it in class and see where I went wrong, if I am wrong, which is more than likely. --[[User:Reckman|Randy Eckman]] 12:19, 5 November 2008 (UTC)
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Yeah, well I must have been really tired because the way I did it I had that <math>(n+1)!<n!</math> which is absolutely ludicrous.  Then I made another mistake after that, but it doesn't matter.  Bell explained it really well and i knew I had seen that problem somewhere else, I just forgot about it. [[User:Jhunsber|His Awesomeness, Josh Hunsberger]]

Latest revision as of 17:45, 5 November 2008

Okay, I'm having a bit of trouble with this one. Neither the root test nor the ratio test is pretty. Should I try to use a different test? It's kind late, so my thoughts are a bit scrambled. I think I'm gonna try to rewrite the sum and see what that gets me.

$ \frac{n!}{n^n}=\frac{(n-1)!}{n^{n-1}} $

Don't know what that will do, but we'll see. His Awesomeness, Josh Hunsberger

Nevermind, I figured it out. You just have to do a second comparison between the resultant denominator in the limit with a simpler version, and show how that the limit is less than one. If you have difficulty, just give me a shout

His Awesomeness, Josh Hunsberger

I got that it diverged because the limit went to infinity...maybe I'll just ask about it in class and see where I went wrong, if I am wrong, which is more than likely. --Randy Eckman 12:19, 5 November 2008 (UTC)

Yeah, well I must have been really tired because the way I did it I had that $ (n+1)!<n! $ which is absolutely ludicrous. Then I made another mistake after that, but it doesn't matter. Bell explained it really well and i knew I had seen that problem somewhere else, I just forgot about it. His Awesomeness, Josh Hunsberger

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