0
$\begingroup$

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.

$\endgroup$
10
  • $\begingroup$ 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
2
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Ehm. Isn't this a normal "sinc" function?;) $\endgroup$ – pisoir Oct 28 '17 at 22:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.