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.
$\endgroup$ 52 Answers
$\begingroup$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.
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ 1
