Determine whether the sequence A(n)=nsinn/n^3 converges or diverges. Eplain why it does or does not converge. It is converges find the limit
To determine whether the sequence A(n) = nsin(n)/n^3 converges or diverges, we need to analyze its behavior as n approaches infinity.
First, let's rewrite the sequence as follows: A(n) = (n*sin(n))/n^3 = (sin(n))/n^2.
To determine convergence, we can use the limit comparison test with a known convergent or divergent sequence. In this case, we can compare it to the sequence B(n) = 1/n^2, which is a known convergent sequence.
Now, let's calculate the limit of the ratio of the two sequences:
L = lim(n→∞) of (A(n)/B(n)) = lim(n→∞) of ((sin(n))/n^2)/(1/n^2) = lim(n→∞) of sin(n).
Since the function sin(n) does not have a limit as n approaches infinity (it oscillates between -1 and 1 infinitely), the limit L does not exist.
Thus, the sequence A(n) = nsin(n)/n^3 diverges because the limit does not exist.
Therefore, we can conclude that the sequence A(n) does not converge.