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$$\sum\limits_{n=1}^{\infty} (-1)^{n-1}\tan\left(\frac{1}{n\sqrt{n}}\right)$$

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 $a_n > a_{n+1}$. 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?

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  • $\begingroup$ You need to be more careful in your proof. If $a_n\sim b_n$ and either sequence remains positive (or negative), then the series $\sum a_n$ and $\sum b_n$ have the same nature. However, we do NOT have $\sum a_n \sim \sum b_n$. Take $\sum\frac{1}{n^2}$ and $\sum\frac{1}{n(n+1)}$ for example. $\endgroup$ – charlus Jul 30 at 14:34
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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.

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  • $\begingroup$ And also it is needed that $\tan$ is $\mathcal C^2$, no? $\endgroup$ – Maximilian Janisch Jul 30 at 17:54
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We need to refer to limit comparison test

$$\lim_{n\to\infty}\frac{\tan\left(\frac1{n\sqrt n}\right)}{\frac1{n\sqrt n}}=1$$

which implies that the absolute series converges since $\sum\limits_{n=1}^{\infty} \frac{1}{n\sqrt{n}}$ converges.

As an alternative, by direct comparison test since for $0\le x \le 1$ we have that $\tan x \le x+x^3$

$$\sum\limits_{n=1}^{\infty}\tan{\frac{1}{n\sqrt{n}}}\le\sum\limits_{n=1}^{\infty} \frac{1}{n\sqrt{n}}+\sum\limits_{n=1}^{\infty} \frac{1}{n^4\sqrt{n}}$$

Since absolute series converges, we can conclude that also the alternating series converges.

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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

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