# Demonstrating Boundedness of Series cos(n): A Challenge

• YAHA
In summary, the conversation is discussing how to show that the series cos(n) for n=0,1,...,inf is bounded. The participants consider using a bound for the individual terms, but then realize they need to show a bound for the entire series. One suggests using e^{in} and the geometric series formula, while another suggests finding the total sum from 1 to infinity. However, it is noted that the series does not converge, but is bounded.
YAHA

## Homework Statement

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

## 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?

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So what's your bound for the individual terms?

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

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

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

How so? Thats not quite obvious to me.

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?

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.

YAHA said:
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 $e^{i n}=\cos n+i \sin n$.

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

YAHA said:
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.

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?

YAHA said:
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.

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.

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.

## 1. What is the series cos(n)?

The series cos(n) is a mathematical series that consists of the cosine function applied to the natural numbers, starting from n = 1. In other words, the series is 1 + cos(2) + cos(3) + cos(4) + ...

## 2. What does it mean to demonstrate boundedness of a series?

Demonstrating boundedness of a series means showing that the series has a finite limit, or in other words, it does not tend to infinity. In simple terms, it means that the values of the series do not become infinitely large or small as n increases.

## 3. Why is demonstrating boundedness of the series cos(n) a challenge?

Demonstrating boundedness of the series cos(n) is a challenge because the values of the cosine function oscillate between -1 and 1 as n increases, making it difficult to determine if the series has a finite limit. Additionally, the series does not converge to a specific value, making it more challenging to show that it is bounded.

## 4. How can one demonstrate boundedness of the series cos(n)?

One way to demonstrate boundedness of the series cos(n) is by using mathematical techniques such as the limit comparison test or the ratio test. These tests involve comparing the series to a known convergent or divergent series and using the results to determine the boundedness of the series cos(n).

## 5. Why is it important to demonstrate boundedness of a series?

Demonstrating boundedness of a series is important because it helps us understand the behavior of the series as n increases. It also allows us to determine if the series will converge or diverge, which has implications in various fields such as physics, engineering, and economics. Additionally, it helps us make predictions and draw conclusions about the series.

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