Does the Sequence a_n = sin(2πn) Converge or Diverge?

Click For Summary

Homework Help Overview

The discussion revolves around the convergence or divergence of the sequence defined by a_{n} = sin(2πn). Participants are exploring the behavior of this sequence as n approaches infinity.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants attempt to evaluate the limit of the sequence and question the apparent contradiction between numerical evaluations and limit behavior. There is discussion about the distinction between limits defined for natural numbers versus real numbers.

Discussion Status

There is an ongoing exploration of the implications of limits in different contexts, with some participants providing insights into the relationship between limits for natural and real numbers. Confusion regarding the convergence of the sequence is acknowledged, but no consensus has been reached.

Contextual Notes

Participants note the potential confusion arising from the definitions of the sine function for different domains and the implications for determining convergence.

Bipolarity
Messages
773
Reaction score
2

Homework Statement



Does the following sequence converge, or diverge?
a_{n} = sin(2πn)

Homework Equations


The Attempt at a Solution



\lim_{n→∞} sin(2πn) does not exist, therefore the sequence should diverge? But it actually converges to 0?

I appreciate all help thanks.

BiP
 
Physics news on Phys.org
Write out the numerical value of a few terms of that sequence.
 
LCKurtz said:
Write out the numerical value of a few terms of that sequence.

They are all 0 ?? So the sequence converges, but how come when we apply the limit it diverges?

BiP
 
I have no idea what you mean by "when we apply the limit it diverges".

\lim_{n=0} sin(2\pi n)= 0, trivially.
 
HallsofIvy said:
I have no idea what you mean by "when we apply the limit it diverges".

\lim_{n=0} sin(2\pi n)= 0, trivially.

I think there is some confusion between sin(2\pi n) defined for natural numbers, and that defined for real numbers. If we take the limit of it as it is defined for natural numbers, it converges to 0 trivially. But if we take the limit of it as it is defined for all real x, its limit does not exist.

Does this mean we cannot take the limit of the function defined for reals to determine the convergence of the function (or sequence) defined for naturals?

BiP
 
Bipolarity said:
I think there is some confusion between sin(2\pi n) defined for natural numbers, and that defined for real numbers. If we take the limit of it as it is defined for natural numbers, it converges to 0 trivially. But if we take the limit of it as it is defined for all real x, its limit does not exist.

Does this mean we cannot take the limit of the function defined for reals to determine the convergence of the function (or sequence) defined for naturals?

BiP

Yes and no. If the limit exists over the reals then it exists over the naturals. The converse is false. Also the limit may exist over the naturals but not over the reals as that example shows.
 
LCKurtz said:
Yes and no. If the limit exists over the reals then it exists over the naturals. The converse is false. Also the limit may exist over the naturals but not over the reals as that example shows.

If the limit over the reals diverges specifically to infinity, must the limit over the naturals also diverge to infinity?

BiP
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 34 ·
2
Replies
34
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K