Studying the convergence of the series $\sum_{n=1}^\infty\sin\frac1{n}\log\left(1+\sin\frac1{n}\right)$ Study the convergence of the series 
$$\sum_{n=1}^\infty\sin\frac1{n}\log\left(1+\sin\frac1{n}\right)$$ 
This is what I came up with
$$\lim_{x\to \infty}\frac{\sin\frac1{n}\log\left(1+\sin\frac1{n}\right)}
{\sin^2\frac1{n}}= 1 $$ 
This implies that 
$$\sin\frac1{n}\log\left(1+\sin\frac1{n}\right) \sim {\sin^2\frac1{n}}$$
using the inequality $\sin{x}\lt x$   $\left(0\le x \lt \pi\right)$
$${\sin^2\frac1{n}} \lt \frac1{n^2}$$
Since  $\sum_{n=1}^\infty\frac1{n^2}$   converges so does $\sum_{n=1}^\infty\sin^2\frac1{n}$ this implies the convergence of $$\sum_{n=1}^\infty\sin\frac1{n}\log\left(1+\sin\frac1{n}\right)$$ 
Is this right?
 A: Thanks to Mark Viola comment, we know $\dfrac{1}{n}<1$ is in first quadrant so $\sin\dfrac{1}{n}>0$, then
$$\sum_{n=1}^\infty\left(\sin\frac{1}{n}\right)\ln\left(1+\sin\frac1{n}\right)<\sum_{n=1}^\infty\left(\sin\frac{1}{n}\right)^2<\sum_{n=1}^\infty\frac{1}{n^2}=\zeta(2)$$
A: Your conclusion is right but we need to clarify some issue.
You have shown that
$$\sin\frac1{n}\log\left(1+\sin\frac1{n}\right) \sim {\sin^2\frac1{n}}$$
and then
$${\sin^2\frac1{n}} \lt \frac1{n^2}$$
thus by comparison test we have that since $\sum_{n=1}^\infty\frac1{n^2}$ also $\sum_{n=1}^\infty\sin^2\frac1{n}$ converges.
Now to conclude we have two choice:


*

*Refer to limit comparison test to show the convergence of the given series given the convergence of $\sum_{n=1}^\infty\sin^2\frac1{n}$

*Refer again to the comparison test showing that (see the comment by Mark Viola)


$$\sin\frac1{n}\log\left(1+\sin\frac1{n}\right)\le \sin \frac1{n^2}\le \frac1{n^2}$$
As a simpler alternative, in my opinion, from here
$$\sin\frac1{n}\log\left(1+\sin\frac1{n}\right)\sim \frac1{n^2}$$
you can directly conclude by limit comparison test with the convergent $\sum_{n=1}^\infty\frac1{n^2}$.
