A question about limsup of the sequence of the average

Question

Let $\{a_n\}$ be a sequence of real non-negative numbers

Define $S_n = \frac{\sum_{i=1}^n a_i}{n}$

Prove that

$$\liminf(a_n) \leq \liminf(S_n) \leq \limsup(S_n) \leq \limsup(a_n)$$

I wanted to show that $\inf\{a_n:n \geq t\} \leq \inf\{S_n:n\geq t\}$ for all $t$

However this doesn't seem to be true

The inequality seems to hold only at n goes to infinity but not every n individually

Can anyone help me?

Answer 1

The statement follows directly by Stolz–Cesàro theorem

$$\liminf \frac{A_{n+1}-A_n}{B_{n+1}-B_n} \leq \liminf \frac{A_{n}}{B_n} \leq \limsup \frac{A_{n}}{B_n} \leq \limsup \frac{A_{n+1}-A_n}{B_{n+1}-B_n}$$

by $A_n=S_n$ and $B_n=n$.

Answer 2

Let $l$ be the $\lim\inf$ of $a_n$. Then for all $\varepsilon > 0$, there exists an $N \in \mathbb{N}$ such that $$n \ge N \implies a_n \ge l - \varepsilon.$$ Let $M = \sum_{k=1}^{N-1} a_k.$ Then, $$n \ge N \implies \sum_{k=1}^n a_n = M + \sum_{k=N}^n a_n \ge M + (n - N + 1)(l - \varepsilon).$$ Therefore, when $n \ge N$, we have $$\frac{\sum_{k=1}^n a_n}{n} \ge \frac{M}{n} + \left(1 - \frac{N - 1}{n}\right)(l - \varepsilon) \to l - \varepsilon.$$ Hence, every subsequence of $\frac{\sum_{k=1}^n a_n}{n}$ must converge to at least $l - \varepsilon$, where $\varepsilon > 0$ is arbitrary. Thus, every subsequence must converge to at least $l$, proving the inequality of the limits inferior.