How to determine whether $\sum_{n = 1}^\infty\frac1n\sin(\frac{1}{n})$ is convergent or divergent

Question

How to determine whether the series converges or diverges: $$\sum_{n = 1}^{\infty}\frac{1}{n}\sin(\frac{1}{n})$$ I think here Limit Comparison Test should be applied but I am not sure how to start.

Answer 1

From the Mean Value Theorem follows that there exist some $y_n\in(0,\tfrac1n)$ such that $$\frac{\sin(\frac1n)-0}{\frac1n}=\cos(y_n)\leq 1$$ then $\sin (\frac1n)\le\frac1n$. And $$\tfrac1n\sin(\tfrac1n)\leq\tfrac1{n^2}$$

So the series is convergent from the comparison test.

Answer 2

Here are two arguments :

1) As $n\to\infty$ :

$$\frac1n\sin\left(\frac1n\right)\sim\frac1{n^2}$$

and we know that the series $\displaystyle{\sum_{n\ge1}\frac1{n^2}}$ is convergent. Hence the convergence of $\displaystyle{\sum_{n\ge1}\frac1n\sin\left(\frac1n\right)}$

2) It is well know that :

$$\forall t\in[0,+\infty),\sin(t)\le t$$

Hence :

$$\forall n\in\mathbb{N^\star},\,0\le\frac1n\sin\left(\frac1n\right)\le\frac1{n^2}$$which gives, by comparison, the conclusion.