# Determine Convergent or Divergent

1. May 6, 2012

### e179285

an = [cos (3n) )] /n.cos(1/n)

My solution is , I wrote an as

liman→∞ [cos (3n) )] /[ cos(1/n) / (1/n) ) .

We get liman→∞ cos(3n) / ( 1/0 ) .

I think solution of this limit is zero but ı'm not sure cos(3n) .I think cos(3n) as a number and number/infinity is zero .As a result of this ı say it is a convergent.

Is the correct approach to the question ?

2. May 6, 2012

### Bohrok

You know that cos n ≤ 1 for any n, so cos(3n) ≤ 1 and cos(3n)/n ≤ 1/n

For the denominator I wouldn't write it as cos(1/n) / (1/n). Try thinking of 1/(n·cos(1/n)) as (1/n)·(1/cos(1/n))

What's the limit of 1/cos(1/n) as n→∞?

3. May 7, 2012

### Ray Vickson

I don't think it is as simple as the OP indicates. Of course, for large n we can consider cos(1/n) to be essentially 1, so we almost have cos(3^n)/n --> 0. However, the sign changes due to cos(3^n) is very important; as n grows there is increasingly rapid oscillation in sign.

RGV

Last edited: May 7, 2012
4. May 7, 2012

### sharks

Use the squeeze theorem on $\frac{\cos(3^n)}{n}$, then evaluate $\frac{1}{\cos(1/n)}$. Using the properties of limits: $\lim (a.b)=\lim a . \lim b$, you should get the answer.

5. May 7, 2012

### Ray Vickson

No, it won't. The series sum 1/n diverges, but the series sum (-1)^n/n converges, so the sign changes are vital. In the current problem there are rapid sign changes (due to oscillation in cos(3^n)), but it is not clear whether the series alternates in any sense (that is, with any kind of nice pattern). I really do believe the problem is non-trivial.

RGV

6. May 7, 2012

### sharks

But the problem involves sequence $a_n$ not series.

7. May 7, 2012

### Ray Vickson

OK, then it's easy. I guess I thought the OP was doing what so many others have done, viz., saying sequence they mean series. I should just have taken him at his word.

RGV