# Homework Help: Sequence Bound

1. Mar 16, 2012

### YAHA

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

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

2. Relevant equations

3. 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?

Last edited: Mar 16, 2012
2. Mar 16, 2012

### Dick

So what's your bound for the individual terms?

3. Mar 16, 2012

### YAHA

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

4. Mar 16, 2012

### Dick

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

5. Mar 16, 2012

### YAHA

How so? Thats not quite obvious to me.

6. Mar 16, 2012

### 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?

7. Mar 16, 2012

### 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.

8. Mar 16, 2012

### Dick

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

9. Mar 16, 2012

### YAHA

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

10. Mar 16, 2012

### Dick

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.

11. Mar 16, 2012

### 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?

12. Mar 16, 2012

### Dick

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

13. Mar 16, 2012

### YAHA

But if you can obtain a finite number for geometric series of e$^{in}$, doesnt it mean that its real part is finite and cos(n) thus converges to a finite number? I think i am misunderstanding something.

14. Mar 16, 2012

### 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.