Please explain what method you used to prove so. $$\sum_{n=3}^\infty \frac{\tan\left(\frac{\pi}{n}\right)}{n}$$
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Convergent, since they approach $\sum \pi/n^2$ as $n$ approaches $\infty$ – Yuriy S Jun 11 '16 at 21:55
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oh ok so you used the comparison test. The thing is that I wasn't sure if I should use that test because I didn't know if the series would always contain positive terms. – tildawn28 Jun 11 '16 at 21:59
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Convergent. The general term behaves like $\frac{1}{n^2}$. – Jack D'Aurizio Jun 11 '16 at 22:04
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Looks like you guys are using $tan(x) \sim x$ – IAmNoOne Jun 11 '16 at 22:05
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2Not sure it contains positive terms? With $0<\frac\pi n\le \frac\pi 3$? – Bernard Jun 11 '16 at 22:25
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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 $\pi^3/3$.
Mark Viola
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Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark – Mark Viola Aug 12 '16 at 01:56