Difference between revisions of "User talk:Cosinator"

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Thanks for contributing to the AoPSWiki.  You might want to look around at a few articles to learn more about standard syntax, or read the [[AoPSWiki tutorial]] (still under contruction, but useful).  Be sure to record the reasons for your edits when you edit an article.--[[User:MCrawford|MCrawford]] 12:31, 20 June 2006 (EDT)
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Thanks for contributing to the AoPSWiki.  You might want to look around at a few articles to learn more about standard syntax, or read the [[AoPSWiki:Tutorial|AoPSWiki tutorial]] (still under contruction, but useful).  Be sure to record the reasons for your edits when you edit an article.--[[User:MCrawford|MCrawford]] 12:31, 20 June 2006 (EDT)
  
 
When you create a new article, please bold the name of the term for which the article is about.  Thanks.--[[User:MCrawford|MCrawford]] 17:09, 20 June 2006 (EDT)
 
When you create a new article, please bold the name of the term for which the article is about.  Thanks.--[[User:MCrawford|MCrawford]] 17:09, 20 June 2006 (EDT)
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In the [[Harmonic series]] article, you wrote:
 
In the [[Harmonic series]] article, you wrote:
 
"
 
"
We could have also shown that the harmonic series diverges by showing that <math>\displaystyle\lim_{n\to\infty}\frac{\frac{1}{n+1}}{\frac{1}{n}}=1</math> and that the limit must be less than 1 for it to converge.
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We could have also shown that the harmonic series diverges by showing that <math>\lim_{n\to\infty}\frac{\frac{1}{n+1}}{\frac{1}{n}}=1</math> and that the limit must be less than 1 for it to converge.
 
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Yes that is true. Thanks for revising the article.  
 
Yes that is true. Thanks for revising the article.  
 
--[[User:Cosinator|cosinator]]
 
--[[User:Cosinator|cosinator]]
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Well <math>\lim_{n\to\infty}\frac{\frac{1}{n+1}}{\frac{1}{n}}=\lim_{n\to\infty} \frac{n}{n+1}</math> then spam L'Hopital on it to get <math>\lim_{n\to\infty}\frac{1}{1}=1</math>... And the absolute value of the limit must be less than 1 to converge... Because if it goes to +-1, then we are adding an infinite amount of almost equal things, and that goes to <math>\infty</math>. Or am I just being stupid? --[[User:1=2|1=2]] 08:12, 6 February 2008 (EST)

Revision as of 08:12, 6 February 2008

Thanks for contributing to the AoPSWiki. You might want to look around at a few articles to learn more about standard syntax, or read the AoPSWiki tutorial (still under contruction, but useful). Be sure to record the reasons for your edits when you edit an article.--MCrawford 12:31, 20 June 2006 (EDT)

When you create a new article, please bold the name of the term for which the article is about. Thanks.--MCrawford 17:09, 20 June 2006 (EDT)

In the Harmonic series article, you wrote: " We could have also shown that the harmonic series diverges by showing that $\lim_{n\to\infty}\frac{\frac{1}{n+1}}{\frac{1}{n}}=1$ and that the limit must be less than 1 for it to converge. "

This is not in fact true: consider, for instance, the series $\sum \frac 1{n^2}$. If the limit of the ratio of consecutive terms is 1, you can't say one way or the other about convergence. --JBL 10:24, 23 August 2006 (EDT)


Yes that is true. Thanks for revising the article. --cosinator

Well $\lim_{n\to\infty}\frac{\frac{1}{n+1}}{\frac{1}{n}}=\lim_{n\to\infty} \frac{n}{n+1}$ then spam L'Hopital on it to get $\lim_{n\to\infty}\frac{1}{1}=1$... And the absolute value of the limit must be less than 1 to converge... Because if it goes to +-1, then we are adding an infinite amount of almost equal things, and that goes to $\infty$. Or am I just being stupid? --1=2 08:12, 6 February 2008 (EST)

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