Determine Convergent or Divergent

[e179285]

a_n = [cos (3ⁿ⁾ )] /n.cos(1/n) My solution is , I wrote a_n as

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

We get lim_an→∞ 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 ?

 

[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→∞?

 

[Ray Vickson]

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→∞?

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

 

[DryRun]

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.

 

[Ray Vickson]

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.

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

 

[DryRun]

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

But the problem involves sequence $a_n$ not series.

 

[Ray Vickson]

But the problem involves sequence $a_n$ not series.

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