Using this fancy list:

enter image description here

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.

  • $\begingroup$ Did you try checking the limit of the terms as $n \to \infty$? $\endgroup$ – user296602 Oct 28 '17 at 22:11
  • $\begingroup$ Why do I have to justify the answer? $\endgroup$ – Kenny Lau Oct 28 '17 at 22:11
  • $\begingroup$ @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. $\endgroup$ – JustHeavy Oct 28 '17 at 22:13
  • $\begingroup$ @DevHeavy Then this is why it's important to ask the question that you have, not copy-paste the book's problem statement.... $\endgroup$ – user296602 Oct 28 '17 at 22:14
  • 1
    $\begingroup$ @DevHeavy two people already gave you the same hint. $\endgroup$ – Kenny Lau Oct 28 '17 at 22:16

Hint $$ \lim_{x\to\infty}x\sin(1/x)= \lim_{u\to 0^+}\frac{\sin u}{u} $$

  • $\begingroup$ Going off of that, since the denominator will be zero, that means the limit DNE and the divergence test says that is divergent, correct? Thank you. $\endgroup$ – JustHeavy Oct 28 '17 at 22:31
  • $\begingroup$ @DevHeavy No, not at all. You should recognize the limit as involving the derivative of $\sin$ at zero. $\endgroup$ – user296602 Oct 28 '17 at 22:47
  • $\begingroup$ @user296602 That limit precedes knowledge of that derivative. It is a fundamental result of elementary calculus. $\endgroup$ – zhw. Oct 29 '17 at 0:10
  • $\begingroup$ @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. $\endgroup$ – 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.

  • $\begingroup$ Ehm. Isn't this a normal "sinc" function?;) $\endgroup$ – pisoir Oct 28 '17 at 22:39

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