posted by kelly on .
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.