May 6, 2016

Homework Help: calc

Posted by kelly on Tuesday, April 17, 2007 at 6:03pm.

determine whethere the sequence a(n)=nsinn/n^3 converges or diverges. Explain why it does or does not converge. If it converges find the limit.

Sin(n) is always between -1 and 1. So, you have:

-1/n^2 < a(n) < 1/n^2

Both the lower bound and the upper bound converge to zero, so the sequence converges to zero.

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