Showing that an inequality is true

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Homework Help Overview

The discussion revolves around proving an inequality related to a nonnegative sequence, specifically showing that the average of a subsequence does not exceed the supremum of the sequence beyond a certain index.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of the supremum and how it relates to the elements of the sequence. There is a focus on understanding why the average of the terms in the sequence cannot exceed the supremum. Some participants question the necessity of showing effort in the proof process.

Discussion Status

Participants are actively engaging with the problem, raising questions about the nature of the sequence and its bounds. Some guidance has been offered regarding the implications of the sequence being convergent or divergent, and the relevance of the arithmetic mean in relation to bounds.

Contextual Notes

There is mention of the sequence being nonnegative and the implications of it being convergent or diverging. The discussion also touches on the flexibility of selecting elements from the sequence, not necessarily being consecutive.

Mr Davis 97
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Homework Statement


Suppose that ##N \in \mathbb{N}## and that ##(s_n)## is a nonnegative sequence. Prove that ##\displaystyle \frac{s_{N+1} + s_{N+2} + \cdots + s_n}{n} \le \sup \{s_n ~:~ n > N \}##

Homework Equations

The Attempt at a Solution


I need help explaining why this is true. Supposedly it is obvious, but I can't quite see it...
 
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You have all ##s_k \le \mathcal{S} :=\sup\{s_n\, : \,n> N\}## for all ##k>N##.
Now count the summands on the left, each smaller than ##\mathcal{S}##, so ##M## many of them are smaller than?
 
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Mr Davis 97 said:

Homework Statement


Suppose that ##N \in \mathbb{N}## and that ##(s_n)## is a nonnegative sequence. Prove that ##\displaystyle \frac{s_{N+1} + s_{N+2} + \cdots + s_n}{n} \le \sup \{s_n ~:~ n > N \}##

Homework Equations

The Attempt at a Solution


I need help explaining why this is true. Supposedly it is obvious, but I can't quite see it...
You have to show some effort. Once you do, you may see it...
 
Well, if s_n is a diverging sequence, the claim is trivially true. Suppose the sequence converges, then it's bounded. Can an arithmetic mean of some n consecutive elements in the sequence exceed that bound? Any bound (hint hint, the smallest bound), for that matter.

The elements can also be picked arbitrarily, they don't have to be consecutive. The claim will still hold.

One can generalise even further and state a similar claim for arbitrary sequences, not just nonnegative ones.
 

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