# calculus – Show that \$frac{1}{2k+ksin{k}}\$ is divergent using direct comparison test

I’m not quite sure if I got how the direct comparison test works.

For a series

$$sum_{k=1}^{n}{frac{1}{2k+ksin{k}}}$$
Is the following correct:

$$frac{1}{2k+ksin{k}} sim frac{1}{3k} le frac{1}{2k+ksin{k}} implies sum_{k=1}^{infty}{frac{1}{2k+ksin{k}}} text{ is divergent. }$$