Convergence of alternating infinite series

Question

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?

Answer 1

For small $x$, both $\sin(x)$ and $\tan(x)$ are close to (i.e., within a factor of 2 of) $x$, so that replacement can almost always be done.

And, of course, $\dfrac1{2n-1}\le \dfrac1{n}$.

Therefore the product is less than twice $\dfrac1{n^{3/2}}$, and the sum of this converges.

Answer 2

Compare $a_n=\dfrac{1}{2n-1}\tan {\dfrac{1}{\sqrt n}}$

with $u_n=\dfrac{1}{n^{\frac{3}{2}}}$

Then we have $\lim\dfrac{a_n}{u_n}=\dfrac{\frac{1}{2n-1}\tan \dfrac{1}{\sqrt n}}{n^{\frac{3}{2}}}=\dfrac{\tan \frac{1}{\sqrt n}}{\frac{1}{\sqrt n}}\times \dfrac{1}{2-\frac{1}{n}}\to \frac{1}{2}(\neq 0)$ as $n\to \infty$

Since $u_n$ converges so the series $\sum a_n$ converges absolutely.