Let $ x_n=(1-\frac{1}{n})\sin (\frac{n\pi}{3})$ for n $\ge $1. denote l=limit inferior and s=limite superior Then

  1. -$\sqrt{3}/2\le l\lt s \le \sqrt{3}/2$

  2. -$1/2\le l\lt s \le 1/2$

  3. $l=-1 $,$ s=1$

  4. $l=0=s=0$

My attempt: I manually plotted the sequence terms for n=1,2,... for few n. And then saw that this sequence has 3 limit points namely -$\sqrt{3}/2$, 0 , $\sqrt{3}/2$. So least one is limit inf , largest of limit point is limit sup. But I need suggestions for more efficient way. Thanks

| cite | improve this question | | | | |

Did you only see that the sequence has three limit points or did you actually prove that it is so?

Hinit: First, simplify the expression $\sin\left(\frac{n\pi}{3}\right)$. Then, it will be much easier to

  1. Find the three subsequences converging to the three convergence points.
  2. Show that these are the only three convergence points.
| cite | improve this answer | | | | |
  • $\begingroup$ I only saw it, by plotting values of this sequence on paper $\endgroup$ – BigBang Feb 5 '15 at 8:05
  • $\begingroup$ Plz give some himt on simplyfying sin term $\endgroup$ – BigBang Feb 5 '15 at 8:07
  • $\begingroup$ @BigBang Calculate the term for $n=1,2,3,4,5$. Can you see a pattern? $\endgroup$ – 5xum Feb 5 '15 at 9:53

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.