# Showing that an inequality is true

Poster has been reminded that showing their work on schoolwork problems is mandatory at the PF

## 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 \}##

## The Attempt at a Solution

I need help explaining why this is true. Supposedly it is obvious, but I can't quite see it...

fresh_42
Mentor
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?

nuuskur and Mr Davis 97
LCKurtz
Homework Helper
Gold Member

## 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 \}##

## 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.