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I was tasked with determining whether the following series:

$$\sum_{n=1}^{\infty} \tan\left(\frac{1}{n}\right) $$

converges.

I tried employing the integral test which failed and produced incalculable integrals. Other methods didn't prove effective also. I was suggested that the Maclauren series might be of use here, but I'm not sure how to employ it.

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Or by limit comparison test with $\sum\frac1n$ since by standard limit for $x\to 0\implies\frac{\tan x}{x}\to 1$ and then

$$\frac{\tan\left(\frac{1}{n}\right)}{\frac1n}\to1$$

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  • $\begingroup$ That's an efficient way to do it. $\endgroup$ – Bak1139 Mar 29 '18 at 13:21
  • $\begingroup$ @Bak1139 Yes indeed limit comparison test avoid to use inequalities which sometimes are difficult to handle. $\endgroup$ – gimusi Mar 29 '18 at 13:23
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We can solve this with the inequality $\tan(x)>x$ for $0<x<\pi/2$ as follows $$\sum_{n=1}^\infty \tan\left(\frac{1}{n}\right)>\sum_{n=1}^\infty\frac{1}{n}$$ And you probably already know that the harmonic series diverges (it can be proven by integral test).

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Hint. Note that for any positive integer $n$, $$\tan\left(\frac{1}{n}\right)> \frac{1}{n}.$$ See Why $x<\tan{x}$ while $0<x<\frac{\pi}{2}$?

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Note that $$\tan \left(\frac{1}{n}\right)\sim_{_{\infty}} \frac{1}{n} $$ As $$\tan \left(\frac{1}{n}\right)=\frac{1}{n}+O\left(n^{-3}\right)$$ Which can be derived from the Maclaurin series expansion of $\tan x$.

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