The nth Term Test for Divergence Practice Test

The series diverges because $\sin(1/n)$ is approximately $1/n$.

The series converges because $\lim_{n\to\infty}\frac{1}{n^3}\sin\left(\frac{1}{n}\right)=0$.

No conclusion can be drawn because $\lim_{n\to\infty}\frac{1}{n^3}\sin\left(\frac{1}{n}\right)=0$.

The series converges because the terms are products.

The series diverges because $\lim_{n\to\infty}\frac{1}{n^3}\sin\left(\frac{1}{n}\right)=0$.