Showing that an inequality is true

[Mr Davis 97]

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...

 

[fresh_42]

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?

 

[LCKurtz]

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...

 

[nuuskur]

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.