Difference between revisions of "User talk:Cosinator"

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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)
  
In the [[Harmonic Series]] article, you wrote:
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In the [[Harmonic series]] article, you wrote:
 
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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.
 
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.

Revision as of 09:24, 23 August 2006

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 $\displaystyle\lim_{n\to\infty}\frac{\frac{1}{n+1}}{\frac{1}{n}}=1$ 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)

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