Determine whether this series converges or not. I need to determinate if the following series converges:
$$\sum_{n=1}^{\infty}\sin{\tan{\frac{1}{n}}}$$
I see that $\sum_{n=1}^{\infty}\tan{\frac{1}{n}}$ obviously diverges, so does $\sum_{n=1}^{\infty}\sin\frac{1}{n}$ but still can't figure how to use the comparison test for the series above.
 A: We have
$$\sin y\geq \frac{2}{\pi}y$$
for $0\leq y\leq \pi/2$ (by concavity) and
$$\tan x\geq x$$
for $0\leq x< \pi/2$. Combining these, we get
$$\sin\tan x\geq \frac{2}{\pi} x$$
for $0\leq x\leq \arctan(\pi/2)\simeq 1.00388$. In particular,

$$\sin \tan\frac{1}{n}\geq\frac{2}{n\pi}\quad\text{for all }n\geq 1$$

and hence the series diverges by the comparison test
A: Using $\sin \epsilon \sim \epsilon$ ant $\tan \epsilon \sim \epsilon$ as $\epsilon\to 0$ and the quotient comparison test:
$$\lim_{n\to +\infty}\dfrac{\sin{\tan{\dfrac{1}{n}}}}{\dfrac{1}{n}}=\lim_{n\to +\infty}\dfrac{{\tan{\dfrac{1}{n}}}}{\dfrac{1}{n}}=\lim_{n\to +\infty}\frac{\dfrac{1}{n}}{\dfrac{1}{n}}=1\ne 0$$ $$\Rightarrow \sum_{n=1}^{\infty}\sin{\tan{\dfrac{1}{n}}}\sim \sum_{n=1}^{\infty}{{\dfrac{1}{n}}}\text{ (divergent).}$$
A: As this is a series with positive terms, you  can use equivalence: $ \sin u\sim_0 u$, so the series $\sin\tan\dfrac1n\sim_\infty\tan \dfrac1n$, and equivalentseries with positive terms both converge or both diverge.
A: The unction $f(x)=\sin{\tan{\dfrac{1}{x}}}$ is continuous, positive and decreasing in $[1,\infty)$, with integral test we have
$$\sum_{n=1}^{\infty}\sin{\tan{\frac{1}{n}}}=\int_1^\infty\sin{\tan{\frac{1}{x}}}dx=\int_1^0\sin{\tan{u}}\frac{-1}{u^2}dx=\int_0^1\frac{\sin{\tan{u}}}{u^2}dx\geq\int_0^1\frac{1}{2u}dx=\infty$$
