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Guys I have to say if $\sum_n\frac{1}{n}\tan\frac{1}{n}$ diverges or not, can you help me and show me how to do it?

3 Answers 3

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Hint

$$\frac{1}{n}\tan\left(\frac{1}{n}\right)\sim \frac{1}{n^2}.$$

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    More crudely ,since $(\tan x)/x\to 1$ as $x\to 0,$ we have $|\tan (1/n)|<2/n $ for all but finitely many $n$.................+1 – 2017-02-01
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Use the comparison test. When proving the derivative of $\sin$, you most likely derived the following inequality:

$$x<\tan(x)$$

for $x\approx0$. From this, it follows that for large $x$, we have

$$\frac1x>\tan\left(\frac1x\right)$$

Thus, we have

$$\sum_{n=1}^\infty\frac1n\tan\left(\frac1n\right)<\sum_{n=1}^\infty\frac1{n^2}$$

Thus, it converges.

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Using the estimates $\sin\frac{1}{n}\leq\frac{1}{n}$ and $\cos\frac{1}{n}\geq\cos1$, for $n\geq1$, we have that the positive series in question is convergent: $$ \sum_{n=1}^\infty\frac{1}{n}\tan\frac{1}{n}\leq\frac{1}{\cos1}\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6\cos1}. $$