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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?

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  • $\begingroup$ 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

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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

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Much less is needed.

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

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