Real Analysis: Prove Upper Bound of Sum of Bounded Sequences

In summary: I don't know. I think you can. I would write it the same way as the one above, but that's only another way to say the same: ##\{s_n+t_n : n \ge k \} \leq S + \{t_n : n \ge k \} \leq S+T##
  • #1
Mr Davis 97
1,462
44

Homework Statement


Suppose that ##( s_n )## and ## (t_n)## are bounded sequences. Given that ##A_k## is an upper bound for ##\{s_n : n \ge k \}## and ##B_k## is an upper bound for ##\{t_n : n \ge k \}## and that ##A_k + B_k## is an upper bound for ##\{s_n + t_n : n \ge k \}##, show that ##\sup \{s_n + t_n : n \ge k \} \le \sup \{s_n : n \ge k \} + \sup \{t_n : n \ge k \}##

Homework Equations

The Attempt at a Solution


Here is what I know. Since ##A_k##, ##B_k## and ##A_k + B_k## are upper bounds, we can conclude that ##\sup \{s_n + t_n : n \ge k \} \le A_k + B_k##, that ##\sup \{s_n : n \ge k \} \le A_k## and that ##\sup \{t_n : n \ge k \} \le B_k##. I think this might be a start, but I am not sure where to go from here...
 
Physics news on Phys.org
  • #2
Mr Davis 97 said:

Homework Statement


Suppose that ##( s_n )## and ## (t_n)## are bounded sequences. Given that ##A_k## is an upper bound for ##\{s_n : n \ge k \}## and ##B_k## is an upper bound for ##\{t_n : n \ge k \}## and that ##A_k + B_k## is an upper bound for ##\{s_n + t_n : n \ge k \}##, show that ##\sup \{s_n + t_n : n \ge k \} \le \sup \{s_n : n \ge k \} + \sup \{t_n : n \ge k \}##

Homework Equations

The Attempt at a Solution


Here is what I know. Since ##A_k##, ##B_k## and ##A_k + B_k## are upper bounds, we can conclude that ##\sup \{s_n + t_n : n \ge k \} \le A_k + B_k##, that ##\sup \{s_n : n \ge k \} \le A_k## and that ##\sup \{t_n : n \ge k \} \le B_k##. I think this might be a start, but I am not sure where to go from here...
Say ##S=\sup \{s_n : n \ge k \}## and ##T=\sup \{t_n : n \ge k \}##. Then ##S+T \le A_k+B_k##. Now is ##S+T## an upper bound for ##\{s_n + t_n : n \ge k \}\,?##
 
  • #3
fresh_42 said:
Say ##S=\sup \{s_n : n \ge k \}## and ##T=\sup \{t_n : n \ge k \}##. Then ##S+T \le A_k+B_k##. Now is ##S+T## an upper bound for ##\{s_n + t_n : n \ge k \}\,?##
For ##S+T## to be an upper bound, wouldn't ##S+T \ge A_k + B_k## have to be the case?
 
  • #4
Mr Davis 97 said:
For ##S+T## to be an upper bound, wouldn't ##S+T \ge A_k + B_k## have to be the case?
No, they are both only upper bounds. Which one is smaller cannot be concluded. The trick is, that ##(s_n+t_n)## is a construction with only one index, ##(s_n) + (t_n)=(s_n)+(t_m)## has two indexes, will say the single sets aren't paired, so the supremums could "happen" at two different places. However, ##S+T \leq A_k + T \leq A_k+B_k##, so what's left is why ##S+T## is an upper bound for the paired set.
 
  • #5
fresh_42 said:
No, they are both only upper bounds. Which one is smaller cannot be concluded. The trick is, that ##(s_n+t_n)## is a construction with only one index, ##(s_n) + (t_n)=(s_n)+(t_m)## has two indexes, will say the single sets aren't paired, so the supremums could "happen" at two different places. However, ##S+T \leq A_k + T \leq A_k+B_k##, so what's left is why ##S+T## is an upper bound for the paired set.
Why can't I just say that ##S = \sup \{s_n : n \ge k \} \ge s_n##, ##T = \sup \{t_n : n \ge k \} \ge t_n##, and thus ##S + T \ge s_n + t_n##, and hence ##S+T## is an upper bound of ##\{s_n+t_n : n \ge k \}##, so hence ##S+T \ge \sup \{s_n+t_n : n \ge k \}##?
 
  • #6
Mr Davis 97 said:
Why can't I just say that ##S = \sup \{s_n : n \ge k \} \ge s_n##, ##T = \sup \{t_n : n \ge k \} \ge t_n##, and thus ##S + T \ge s_n + t_n##, and hence ##S+T## is an upper bound of ##\{s_n+t_n : n \ge k \}##, so hence ##S+T \ge \sup \{s_n+t_n : n \ge k \}##?
I don't know. I think you can. I would write it the same way as the one above, but that's only another way to say the same: ##\{s_n+t_n : n \ge k \} \leq S + \{t_n : n \ge k \} \leq S+T##
 
  • #7
I think this is a nice case to consider: ## s_n =(-1)^n ; t_n=(-1)^{n+1} ##. EDIT: It may be a nice exercise to determine conditions when the Sup of the sum is the sum of the Sups.
 

1. What is Real Analysis and why is it important?

Real Analysis is a branch of mathematics that deals with the study of real numbers and their properties. It is important because it provides a rigorous foundation for calculus and other areas of mathematics, and it is also widely used in applications such as physics, engineering, and economics.

2. What does it mean to prove an upper bound of a sum of bounded sequences?

Proving an upper bound of a sum of bounded sequences means to show that the sum of the terms in the sequence is always less than or equal to a certain value, known as the upper bound. This is useful in determining the limit of a sequence, as well as in proving the convergence or divergence of a series.

3. How do you prove the upper bound of a sum of bounded sequences?

To prove the upper bound of a sum of bounded sequences, you must first show that the individual terms in the sequence are bounded, meaning that they are always less than or equal to a certain value. Then, you can use mathematical techniques such as induction or the triangle inequality to prove that the sum of the terms is also bounded by a certain value.

4. Can you provide an example of proving the upper bound of a sum of bounded sequences?

Sure, let's say we have the sequence a_n = (-1)^n/n, which is bounded by -1 and 1. We want to prove that the sum of the first n terms in this sequence is bounded by 1. We can do this by using mathematical induction to show that the sum of the first n terms is always less than or equal to 1.

5. Why is it important to prove the upper bound of a sum of bounded sequences?

Proving the upper bound of a sum of bounded sequences is important because it helps us understand the behavior of a sequence and its limit. It also allows us to determine if a series is convergent or divergent, which has important implications in many fields of mathematics and science. Additionally, proving the upper bound of a sum of bounded sequences helps to strengthen our understanding of mathematical concepts and techniques.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
806
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
981
  • Calculus and Beyond Homework Help
Replies
7
Views
644
Replies
4
Views
1K
Back
Top