Is this series divergent or convergent?

Question

Please explain what method you used to prove so. $$\sum_{n=3}^\infty \frac{\tan\left(\frac{\pi}{n}\right)}{n}$$

Answer 1

Note that $\tan(x)\sim x$ as $x\to 0$.

Answer 2

In THIS ANSWER, I showed using elementary inequalities from geometry that the tangent function satisfies the inequalities

$$x\le \tan(x)\le x\sec(x) \tag 1$$

for $0\le x<\pi/2$. Therefore, using $(1)$ we find for $n\ge 3$

$$\left|\frac{\tan\left(\frac{\pi}{n}\right)}{n}\right|\le \frac{\pi}{n^2}\sec(\pi/n)\le \frac{2\pi }{n^2}$$

since the secant function on $[0,\pi/3]$ attains its maximum at $\pi/3$

Finally, using $\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}{6}$, we see that the series of interest converges and is in fact, less than $\pi^3/3$.