Why do I have to use L'Hopital in limit comparison test for $\sum_{n=1}^{\infty} \sin\left(\frac{1}{n}\right)$

I'm trying to determine whether the following series diverges or converges:

$$\sum_{n=1}^{\infty} \sin\left(\frac{1}{n}\right)$$

I use the limit comparison test, comparing it to $$\frac{1}{n}$$, which we know diverges.

$$a_n = \sin\left(\frac{1}{n}\right)$$

$$b_n = \frac{1}{n}$$

Thus using the limit comparison test we do:

$$\lim_{n\to\infty} \frac{\sin\left(\frac{1}{n}\right)}{\frac{1}{n}}$$

When evaluating this limit, why can't I make it of the form:

$$\lim_{n\to\infty} n \sin\left(\frac{1}{n}\right)$$

Isn't $$0 \cdot \infty = 0$$ valid? It turns out not because the correct answer of this limit is $$1$$. Why is it not valid to do it the way I did it?

• No, you can't write $0\times\infty=0$ because $0\times\infty$ is an indeterminate form (for example, consider $\lim_{\to\infty}n^a n^{-b}$ for $a,\,b>0$). See also en.wikipedia.org/wiki/Indeterminate_form – J.G. Dec 9 '18 at 21:59
• It would be a big mistake, because the limit of $n\sin\frac{1}{n}$ is $1$. A form such as $\infty\cdot a$, with $\color{red}{a\ne0}$, gives $\infty$ or $-\infty$ according to $a>0$ or $a<0$. But $\infty\cdot0$ cannot be treated in a simple way. – egreg Dec 9 '18 at 22:36

Indeed we don't need to use l'Hopital at all, simply take $$x=\frac 1 n \to 0$$ to obtain
$$\frac{a_n}{b_n}=\frac{ \sin\left(\frac{1}{n}\right)}{\frac1n}=\frac{\sin x}{x}$$
It suffices to know that $$\sin x>\frac12x$$ for $$0, say.