Q:

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

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.

A:

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.

