Are we to understand that the series in question is the following? $$\sum_{n = 1}^\infty a_n \qquad \text{where} \, a_n = {\sqrt{n}\cos n \over n+9}$$

In terms of absolute convergence, we can say that $\sum a_n$ converges if $\sum |a_n|$ converges. But we have a problem in proving that.

$$0 \lt \left|{\sqrt{n}\cos n \over n+9}\right| \lt {\sqrt{n} \over n+9} \lt {\sqrt{n} \over n} = {1 \over \sqrt{n}} = b_n$$

But the series $\sum b_n$ is not convergent anyway.

We might like to use a comparison with a series like $\sum {1 \over n}$ to prove divergence, but then the fact that some of the terms of the original sequence are negative and some are probably much smaller than ${1 \over n}$ scuppers that plan.

The ${1 \over n^{3 \over 2} + 9}$ in the original post is inaccurate.

Wolfram Alpha suggests that the series might converge. And, being an approximately alternating series with terms tending to zero, it is likely to converge.