# Convergence of a sequence

1. Nov 8, 2012

### Bipolarity

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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

2. Nov 8, 2012

### LCKurtz

Write out the numerical value of a few terms of that sequence.

3. Nov 9, 2012

### Bipolarity

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

BiP

4. Nov 9, 2012

### HallsofIvy

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

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

5. Nov 9, 2012

### Bipolarity

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

6. Nov 9, 2012

### LCKurtz

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.

7. Nov 9, 2012

### Bipolarity

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

BiP