# Series convergence $(-1)^n n\tan\frac{1}{n}$

I tried to prove the convergence of this series with the ratio and root test but neither worked. I also couldn't find another series to compare. I understand that $$\tan\frac{1}{n}$$ is decreasing and the partial sums too.

I have no idea what to try now

$$\sum _{n=1}^{\infty }(-1)^n n\tan\frac{1}{n}.$$

• Hint: near $0$, $\tan(x)$ is very much like $x$ (look at the Taylor series), so your sum is essentially $\sum (-1)^n\cdot 1$, which does not converge... – paul garrett Mar 31 at 16:19

$$n\tan\left(\dfrac1n\right)=\frac1{\cos\left(\dfrac1n\right)}\frac{\sin\left(\dfrac1n\right)}{\dfrac 1n}$$ shows that the general term does not tend to zero.

• When I take the $\underset{n\to \infty }{\text{lim}}$ the right side became $\left( \frac{1}{1}\right) \frac{0}{0}$, because $\frac{0}{0}$ I apply L'Hôpital then it goes to 1. Is it right? – Carlos Eduardo Mar 31 at 16:46
• @CarlosEduardo Are you familiar with the classical limit $$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1?$$ – Gary Mar 31 at 16:53
• @CarlosEduardo that is one way of doing it, yes. For another way (using Maclaurin series of $\sin(x)$), see my answer. – Adam Rubinson Mar 31 at 16:53
• @Gary yes. I didn't notice I could use the it. – Carlos Eduardo Mar 31 at 17:01

Expanding on Yves Daoust's terse answer,

$$\displaystyle\lim_{n\to\infty}\cos\left(\dfrac1n\right) = 1\neq 0,\$$ so $$\ \displaystyle\lim_{n\to\infty}\left(\frac1{\cos\left(\dfrac1n\right)}\right) = \frac{1}{1} = 1,\$$ and here, we see that:

$$\displaystyle\lim_{n\to\infty}\left(\frac{\sin\left(\dfrac1n\right)}{\dfrac 1n}\right) = \displaystyle\lim_{n\to\infty}\left(\frac{\left(\dfrac1n\right)-\frac{\left(\dfrac1n\right)^3}{3} + \frac{\left(\dfrac1n\right)^5}{5}+\ldots}{\dfrac 1n}\right) = \displaystyle\lim_{n\to\infty}\left(1-\frac{\left(\dfrac1n\right)^2}{3} + \frac{\left(\dfrac1n\right)^4}{5}+\ldots\right) = 1.$$

Therefore,

$$\displaystyle\lim_{n\to\infty}\left(n\tan\left(\dfrac1n\right)\right)=\displaystyle\lim_{n\to\infty}\left(\frac1{\cos\left(\dfrac1n\right)}\frac{\sin\left(\dfrac1n\right)}{\dfrac 1n}\right) = \displaystyle\lim_{n\to\infty}\left(\frac1{\cos\left(\dfrac1n\right)}\right)\ \cdot\ \displaystyle\lim_{n\to\infty}\left(\frac{\sin\left(\dfrac1n\right)}{\dfrac 1n}\right) = 1\cdot 1 = 1.$$