# Is $\sum_{n=1}^\infty {n\sin({1\over n})}$ divergent or convergent? Justify the answer.

Using this fancy list:

I think I can rule out the options 1 through 5, but I'm not sure. The book I have gotten this problem from does not have any examples that use a trig function. I think I'll need to use option 6, what do you guys think? Or should I use a comparison test?

Thank you for any help, I'm really bad at recognizing these series and what I should use to solve them.

• Did you try checking the limit of the terms as $n \to \infty$? – user296602 Oct 28 '17 at 22:11
• Why do I have to justify the answer? – Kenny Lau Oct 28 '17 at 22:11
• @KennyLau Sorry if it seems subjective, I did not mean to word it that way. It is just how it is worded in the book. I think it means just to show how to figure out if it is convergent or divergent. – JustHeavy Oct 28 '17 at 22:13
• @DevHeavy Then this is why it's important to ask the question that you have, not copy-paste the book's problem statement.... – user296602 Oct 28 '17 at 22:14
• @DevHeavy two people already gave you the same hint. – Kenny Lau Oct 28 '17 at 22:16

Hint $$\lim_{x\to\infty}x\sin(1/x)= \lim_{u\to 0^+}\frac{\sin u}{u}$$
• @DevHeavy No, not at all. You should recognize the limit as involving the derivative of $\sin$ at zero. – user296602 Oct 28 '17 at 22:47
• @zhw. The asker's comment implies that they have the belief that $\lim_{u \to 0} \frac{\sin u}{u}$ does not exist because of the denominator. I was trying to point out that this is incorrect. – user296602 Oct 29 '17 at 0:21
You should consider the limit $$\lim_{x\to\infty} xsin(\frac{1}{x})$$ It is equivalent to another you limit you may have seen before. $$\lim_{x\to 0^+} \frac{sin(x)}{x}$$ It should be obvious at this point that you can use the divergence test to show that the sum diverges.