Answer
Diverges
Work Step by Step
Let $u_n= \tan (\dfrac{1}{n})$ and $v_n=\dfrac{1}{ n}$
Now, $\lim\limits_{n \to \infty}\dfrac{u_n}{v_n} =\lim\limits_{n \to \infty}\dfrac{ \tan (\dfrac{1}{n})}{1/n}=\dfrac{\infty}{\infty}$
Need to apply L-Hospital's rule.
$\implies \lim\limits_{n \to \infty} \dfrac{\sec^2 (\dfrac{1}{n})}{1}= \sec (0)=1 \ne 0 \ne \infty $
Here, $\Sigma_{n=1}^\infty \dfrac{1}{n^{(1)}}$ is a divergent $p$-series test with $p \leq 1$
Hence, the series diverges by the limit comparison test.