## Calculus (3rd Edition)

The series $\Sigma_{n=1}^{\infty} \frac{n-\cos n}{n^3}$ converges.
Using the limit comparison test with $b_n=\frac{1}{n^2}$, we have $$L=\lim_{n\to \infty} \frac{a_n}{b_n}= \lim_{n\to \infty} \frac{(n-\cos n)/n^3}{1/n^2}\\ =\lim_{n\to \infty} \frac{n-\cos n}{n}=\lim_{n\to \infty} (1-\frac{ \cos n}{n})=1$$ where we used the fact that $\lim_{n\to \infty} \frac{ \cos n}{n}=0$. Since $\Sigma_{n=1}^{\infty}\frac{1}{n^2}$ is a convergent p-series and $L\gt 1$ then the series $\Sigma_{n=1}^{\infty} \frac{n-\cos n}{n^3}$ converges.