Divergence of series sin(1/n)

Question

Since $-1 \le \sin(1/n)\le 1$

and

$\lim_{n\to \infty} -1$ $\neq$ $\lim_{n\to \infty} 1$

can I use the nth-term test to prove that the series will diverge?

I've only seen the problem done using the limit comparison test and am not sure if I can use the nth-term test.

Answer 1

This is not valid. Indeed, note that

$$\lim_{n\to\infty}\sin(1/n)=0$$

So it passes the $n$th term test. However, it does diverge. Note that:

$$n\ge1\implies\sin(1/n)\ge\frac{\sin(1)}n$$

So we may use the direct comparison test.

Answer 2

For $0\le x\le \pi/2$, the sine function is bounded by

$$\frac2\pi x\le \sin(x)\le x$$

Therefore, we assert that

$$\sum_{n=1}^N \sin\left(\frac1n\right) \ge \frac2\pi \sum_{n=1}^N \frac1n$$

And hence, by comparison with the harmonic series, the series $\sum_{n=1}^\infty \sin\left(\frac1n\right)$ diverges.