Homework Help: Partial limits of cos(pi*n/3)

1. Nov 10, 2012

peripatein

Hi,
Trying to find all partial limits of cos(pi*n/3), I separated it into:
a_3k -> -1
a_6k -> 1

Is this a valid approach? Are there any other partial limits?

2. Nov 10, 2012

HallsofIvy

Essentially you are looking at n= 3 (mod 6) and n= 0 (mod 6), which is a very good idea, but those two are not the only possibilities. Suppose n is equal 1 (mod 6). That is, n= 1+ 6k. In that case, $\pi n/3= \pi/3+ 2\pi k$. Now, $cos(A+ B)= cos(A)cos(B)- sin(A)sin(B)$ so $cos(\pi/3+ 2\pi k)= cos(\pi/3)cos(2\pi k)+ sun(\pi/3)sin(2\pi k)= cos(\pi/3)$ for all k.

Similarly, look at n= 2+ 6k, n= 4+ 6k, and n= 5+ 6k.

Last edited by a moderator: Nov 11, 2012
3. Nov 10, 2012

peripatein

Thanks! Hence this sequence has only four partial limits. Is that correct?

4. Nov 10, 2012

peripatein

And what about the sequence ncos(pi*n/4). I believe it has only two partial limits, converging to +-infinity. Is that true?

5. Nov 10, 2012

haruspex

In all sequences cos(n*pi/m), constant integer m, each value that the sequence can take occurs infinitely often. So every value is the limit of a subsequence. So, yes, for m=3 there are 4 values.
For n*cos(n*pi/4), what are the possible values in the sequence?

6. Nov 10, 2012

peripatein

Would it be incorrect to say that n*cos(n*pi/4) has merely two partial limits (+-infinity) as cos(n*pi/4) is bounded whereas n isn't?

7. Nov 10, 2012

haruspex

I could be wrong, but I wouldn't have thought infinity qualified as the limit of a sequence. It isn't a number, so you can't converge to it.
OTOH, there is one finite limit for this sequence. What are all the possible values of cos(n*pi/4) ?

8. Nov 10, 2012

peripatein

Infinity, plus and minus, could well serve as PL's in this case, as thus the question is formulated in the textbook. In any case, the possible values are +-1/sqrt(2), 0 and 1. But that multiplied by infinity would either be infinity or, in the case of 0, undetermined, would it not?

9. Nov 11, 2012

haruspex

No, you're not multiplying cos(n*pi/4) by infinity, you're taking a limit. In particular, what is the limit of n*cos(n*pi/4) for the subsequence n = 4k+2?

10. Nov 11, 2012

peripatein

I still don't see why it wouldn't be infinity. Could you help me see it?

11. Nov 11, 2012

haruspex

(4k+2)cos((4k+2)pi/4) = (4k+2)cos(pi*k+pi/2) = 0 for all k. Therefore the sequence converges to 0.

12. Nov 11, 2012

peripatein

Thank you. Are there are any other partial limits?

13. Nov 11, 2012

vela

Staff Emeritus
You tell us.