Let $(a_n)_{n\in\mathbb{N}}$ be a bounded sequence and let $B_n = \frac{1}{n} \sum_{i=1}^n a_i$ for each $n \in \mathbb{N}$. Prove that $\liminf a_n \le \liminf B_n \le \limsup B_n \le \limsup a_n$.
Please check whether the proof that i have mentioned below is correct or not.If there is a better or correct proof can you please show it here .
Since $(a_n)$ is bounded, $\limsup a_n$ exists as a real number. Let $L = \limsup a_n$ for some $L \in \mathbb{R}$.
Let $ε>0$. Then $\exists K \in \mathbb{N}$ such that for each $n>K$ , a_n < L+ε$.
Now let $n \in \mathbb{N}$. Then
$a_{K+1} < L+ε$.
$a_{K+2} < L+ε$.
$\cdots$
$a_{K+n} < L+ε$.
Thus $\frac{1}{n} ( a_{K+1} + a_{K+2} + \cdots + a_{K+n} ) < L+ε$.
Hence $\limsup \frac{1}{n} ( a_{K+1} + a_{K+2} + \cdots + a_{K+n} ) \le L+ε$.
Thus $\limsup \frac{1}{n} ( a_{1} + a_{2} + \cdots + a_{n} ) \le L$.
Your proof is wrong because $\frac{1}{n} ( a_{K+1} + a_{K+2} + a_{K+3} + \cdots + a_{K+n} )$ is not an element of your sequence $(B_n)_{n\in\mathbb{N}}$.
Hint
Given any $ε > 0$, let $K$ be as you stated. Then as $n \to \infty$, we have $n > k$ and hence $B_n = \frac{1}{n} \sum_{i=1}^n a_i = \frac{1}{n} ( \sum_{i=1}^K a_i + \sum_{i=k+1}^n a_i ) \le \cdots \text{[use the property of $K$ here]} \cdots \to \cdots$. Thus as $n \to \infty$ eventually $B_n < L+2ε$. Since this works for any such $ε$, we get $\limsup B_n \le L$.