Studying the convergence of the series $\sum_{n=1}^\infty\sin\frac1{n}\log\left(1+\sin\frac1{n}\right)$

Question

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?

Answer 1

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:

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

Answer 2

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