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

• 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. – charlus Jul 30 at 14:34

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.

• And also it is needed that $\tan$ is $\mathcal C^2$, no? – Maximilian Janisch Jul 30 at 17:54

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.

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