zeion Posted March 23, 2010 Report Share Posted March 23, 2010 Hi can I get some help on this The problem statement, all variables and given/known data Determine whether the series converges or diverges. [math] \sum \frac{lnk}{k^3} [/math] The attempt at a solution Since lnk always less than 0, so [math] \frac{lnk}{k^3} \leq \frac{1}{k^3}[/math] and [math]\frac{1}{k^3} [/math]diverges so[math] \frac{lnk}{k^3}[/math] diverges. Quote Link to comment Share on other sites More sharing options...

lawcat Posted March 23, 2010 Report Share Posted March 23, 2010 The divergence test is done by taking the limit of the main term. In your case, you have to take the limit of [ ln K / K^3.] To do that , you have to apply L'Hospital's rule, and differentiate numerator and denominator. In that case, you will get the limit of [1/3K^3]. If that limit is 0, then the series is converging. If not, then it diverges. Quote Link to comment Share on other sites More sharing options...

A23 Posted May 14, 2010 Report Share Posted May 14, 2010 The criterion above is not sufficient, for your case it is ok, but for example the series :[math]\sum_k\frac{1}{k}[/math] diverges even if the term tends towards 0. You have comparison criterion or integrating instead of summing, but i do not have here the condition of applicability. Quote Link to comment Share on other sites More sharing options...

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