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?

  • $\begingroup$ 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 $\endgroup$
    – J.G.
    Dec 9, 2018 at 21:59
  • $\begingroup$ 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. $\endgroup$
    – egreg
    Dec 9, 2018 at 22:36

2 Answers 2


Indeed we don't need to use l'Hopital at all, simply take $x=\frac 1 n \to 0$ to obtain

$$\frac{a_n}{b_n}=\frac{ \sin\left(\frac{1}{n}\right)}{\frac1n}=\frac{\sin x}{x}$$

and refer to standard limits.

You shouldn't have doubts about that if you are dealing now with series, in case refer to the related


Much less is needed.

It suffices to know that $\sin x>\frac12x$ for $0<x<\frac \pi 2$, say.


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