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.
To determine whether the sequence a(n) = nsin(n)/n^3 converges or diverges, we can use the squeeze theorem. The squeeze theorem states that if the terms of a sequence are bounded between two other sequences that converge to the same limit, then the original sequence also converges to that limit.
In this case, we have:
-1/n^2 < a(n) < 1/n^2
Both the lower bound -1/n^2 and the upper bound 1/n^2 converge to zero as n approaches infinity. Since the sequence a(n) is trapped between these two converging sequences, we can conclude that a(n) also converges to zero.
Thus, the sequence a(n) = nsin(n)/n^3 converges to zero.
Please note that this explanation relies on the facts that sin(n) is always between -1 and 1, and that the limit of 1/n^2 as n approaches infinity is zero.