Demonstrating Boundedness of Series cos(n): A Challenge

[YAHA]

Homework Statement

I need to show that series cos(n) for n=0,1,...,inf is bounded.

Homework Equations

The Attempt at a Solution

I understand that each of the terms of the sequence is bounded by 1. However, I cannot show the bound for the entire sequence? Could someone give some hints?

 

[Dick]

So what's your bound for the individual terms?

 

[YAHA]

Well ,obviously, -1 <= cos(n) <= 1, so |cos(n)| <=1

 

[Dick]

Well ,obviously, -1 <= cos(n) <= 1, so |cos(n)| <=1

That's a bound for the sequence as well, isn't it?

 

[YAHA]

How so? Thats not quite obvious to me.

 

[Dick]

If you call the general term of the sequence $a_n=\cos n$ then $|a_n| \le 1$. Isn't that what you mean by a sequence bound?

 

[YAHA]

Actually, I meant to say series, not sequence. I need to show bound of the series. My apologies.

Thus, for some n=N, the partial sum is bounded by N. I am struggling to show that the entire sum is bounded.

 

[Dick]

Actually, I meant to say series, not sequence. I need to show bound of the series. My apologies.

Thus, for some n=N, the partial sum is bounded by N. I am struggling to show that the entire sum is bounded.

That's different! Try using $e^{i n}=\cos n+i \sin n$.

 

[YAHA]

Do I need to express cos(n) as e$^{in}$ and then use geometric series?

 

[Dick]

Do I need to express cos(n) as e$^{in}$ and then use geometric series?

Find a partial sum of the geometric series $e^{in}$. Then to find the partial sum of cos(n) just find the real part of that.

 

[YAHA]

I undestand. But isn't it easier to just find the total sum from 1 to infinity of the geometric series and essentially say that both cos(n) and sin(n) converge to the obtained number's real and imaginary parts respectively?

 

[Dick]

I undestand. But isn't it easier to just find the total sum from 1 to infinity of the geometric series and essentially say that both cos(n) and sin(n) converge to the obtained number's real and imaginary parts respectively?

The series doesn't converge. But it is bounded.

 

[YAHA]

But if you can obtain a finite number for geometric series of e$^{in}$, doesn't it mean that its real part is finite and cos(n) thus converges to a finite number? I think i am misunderstanding something.

 

[Dick]

The geometric series formula gives you (1-r^(n+1))/(1-r) for a partial sum. The r^(n+1) part doesn't converge. You have to try to bound the partial sum.