Here is the sum:
$\sum_{n=1}^{\infty}(-1)^n(\frac{1}{2n-1})\tan (\frac{1}{\sqrt{n}})$
Test for absolute convergence, conditional convergence, or divergence.
I used the alternating series test and found that it converges. How do I go about testing for absolute convergence? I tried comparing it to $\sum\frac{1}{n}$ with the Limit Comparison Test but the limit got really hairy when doing L'Hopital's rule. I tried comparing $\sum\tan(\frac{1}{\sqrt{n}})$ to $\sum\frac{1}{\sqrt{n}}$ but I don't know which is bigger. Any thoughts?