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?
Best Answer
Thanks to Mark Viola comment, we know $\dfrac{1}{n}<1$ is in first quadrant so $\sin\dfrac{1}{n}>0$, then $$\sum_{n=1}^\infty\left(\sin\frac{1}{n}\right)\ln\left(1+\sin\frac1{n}\right)<\sum_{n=1}^\infty\left(\sin\frac{1}{n}\right)^2<\sum_{n=1}^\infty\frac{1}{n^2}=\zeta(2)$$