Proof of Absolute Convergence of $\sum\limits_{n=1}^{\infty} (-1)^{n-1}\tan\left(\frac{1}{n\sqrt{n}}\right)$ $$\sum\limits_{n=1}^{\infty} (-1)^{n-1}\tan\left(\frac{1}{n\sqrt{n}}\right)$$
My Attempt: To prove absolute convergence, we must consider $\sum\limits_{n=1}^{\infty}a_n$ and $\sum\limits_{n=1}^{\infty}|a_n|$.
I know that as $\theta \to0$ we can approximate $\tan\theta \sim \theta$. Hence $\sum\limits_{n=1}^{\infty}|a_n|$ becomes:
$$\sum\limits_{n=1}^{\infty}\left|\tan{\frac{1}{n\sqrt{n}}}\right| \sim_{\infty} \sum\limits_{n=1}^{\infty} \frac{1}{n\sqrt{n}}, 0 \leq a_n$$
Which converges by the $p$ test. As noted, the $\theta \to 0$ which means that $a_n &gt; a_{n+1}$. Furthermore:
$$\lim_{n \to \infty} \tan{\frac{1}{n\sqrt{n}}} = 0$$
By the alternating series test, it converges. Since the sums for $a_n$ and $|a_{n}
|$ are both converging, it is absolute convergence by definition. Is this approach correct?
Almost. Just saying that $\tan\theta\sim\theta$ is a bit vague though. I suggest that you add$$\lim_{n\to\infty}\frac{\tan\left(\frac1{n\sqrt n}\right)}{\frac1{n\sqrt n}}=1$$to your proof, which is something that follows from that fact that $\tan0=0$ and that $\tan'(0)=1$. In other words, use the comparison test.
we have  $$n\geq1 , |\tan(\frac{1}{n\sqrt{n}})|\leq|\frac{3}{n\sqrt{n}}|   \implies \sum_{n=1}^{\infty}|\tan(\frac{1}{n\sqrt{n}})(-1)^{n-1}|\leq \sum_{n=1}^{\infty}|\frac{3}{n\sqrt{n}}|$$
and
$$\sum_{n\geq 1}^{}|\frac{3}{n\sqrt{n}}| $$  is convergent $$\implies \sum_{n\geq 1}^{}|\tan(\frac{1}{n\sqrt{n}})(-1)^{n-1}|$$ is convergent $$\implies \sum_{n\geq}{}(-1)^{n-1}\tan(\frac{1}{n\sqrt{n}})$$ Absolutely converges