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
$$\sum _{n=1}^{\infty }(-1)^n n\tan\frac{1}{n}.$$
 A: $$n\tan\left(\dfrac1n\right)=\frac1{\cos\left(\dfrac1n\right)}\frac{\sin\left(\dfrac1n\right)}{\dfrac 1n}$$ shows that the general term does not tend to zero.
A: 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,
$$\displaystyle\lim_{n\to\infty}\left(n\tan\left(\dfrac1n\right)\right)=\displaystyle\lim_{n\to\infty}\left(\frac1{\cos\left(\dfrac1n\right)}\frac{\sin\left(\dfrac1n\right)}{\dfrac 1n}\right) = \displaystyle\lim_{n\to\infty}\left(\frac1{\cos\left(\dfrac1n\right)}\right)\ \cdot\ \displaystyle\lim_{n\to\infty}\left(\frac{\sin\left(\dfrac1n\right)}{\dfrac 1n}\right) = 1\cdot 1 = 1.$$
