If $(a_{n})_{n}$ is a bounded sequence, show that $\liminf_{n\to \infty}a_{n}\leq \liminf_{n\to \infty}\frac{a_{1}+a_{2}+\cdots +a_{n}}{n}$.

Question

I am very lost at this proof. I tried to break it a part but it made it more confusing. I am not sure what algebra or theorems I must use.

Answer 1

We have $\liminf_{n}a_n = \sup_{m \geq 1}\inf_{n\geq m}a_n$

Suppose that $\liminf _n \dfrac{a_1+a_2+\cdots +a_n}{n} = a$. Then for all $m$ $$\liminf _n \dfrac{a_m+a_{m+1}+\cdots +a_n}{n} = a.$$ Notice that $\dfrac{a_m+a_{m+1}+\cdots +a_n}{n}\geq (n-m+1)\frac{\min(a_m,a_{m+1},\ldots, a_n)}{n}$, taking $\liminf$ we get $$a\geq \inf_{n\geq m}a_n$$ Now, taking $\sup$ over all $m$ we obtain $$a\geq \sup_{m \geq 1}\inf_{n\geq m}a_m=\liminf_{n}a_n$$

Answer 2

There is also another way, a little bit more general (like stolz-Cesàro theorem), my attempt follows

Notation: $\Delta y_k= y_{k+1}-y_k$.

Theorem:

Suppose that $(y_k)$ is a crescent sequence with $\lim y_k= \infty$ and $(x_k)$ a bounded sequence. Then $$\lim \inf \frac{\Delta x_k }{\Delta y_k}\leq \lim \inf \frac{x_k}{y_k}. $$

Proof:

Define $\alpha:=\lim \inf \frac{\Delta x_k }{\Delta y_k} $, take any $b$ with $b< \alpha$. There exists $n_0$ so that for $k \geq n_0$ we have $$b \leq \frac{\Delta x_k}{\Delta y_k}. $$ $(y_k)$ is crescent, then $\Delta y_k >0$, multiply both sides $$b \Delta y_k \leq \Delta x_k $$ Apply the sum $\sum_{k=n_0}^{n-1}$ on both sides and use telescopic sum, $$b(y_n-y_{n_0}) \leq x_n-x_{n_0} \Rightarrow b(y_n-y_{n_0})+x_{n_0} \leq x_n$$ $y_n$ is positive for $n$ large enough and $y_n \rightarrow \infty$, so $$b(1-\frac{y_{n_0}}{y_n})+\frac{x_{n_0}}{y_n} \leq \frac{x_n}{y_n} $$ then $$ b \leq\lim \inf \frac{x_n}{y_n}, $$ $b<\alpha=\lim \inf \frac{\Delta x_k }{\Delta y_k}$ was taken as a arbitrary number less then $ \alpha$ then $$\lim \inf \frac{\Delta x_k }{\Delta y_k} \leq \lim \inf \frac{x_n}{y_n}. $$

Example: With $y_n=n$ and $x_k=\sum^{k}_{s=1}a_s$ follows $$ \lim \inf a_k \leq \lim \inf \frac{\sum^{k}_{s=1}a_s}{k}. $$