Consider this well known result:
${\bf Theorem:}$ Let $(a_n)_n\subset\mathbb{R}$ be a sequence and let
$$b_n = \frac{1}{n}\sum_{k=1}^n a_k.$$
Thus we have
$$a_n\to l\implies b_n \to l,$$ (but, in general, the converse does not hold).
Starting from this result, I would like to show that if $a_n$ and $b_n$ are as above, it holds $$\liminf_{n\to +\infty} a_n\leq\liminf_{n\to +\infty} b_n\leq\limsup_{n\to +\infty} b_n\leq\limsup_{n\to +\infty} a_n.$$
I tried in this way:
$l$ belongs to the adherence values of $a_n$, thus $a_{n_k}$ exists such that $a_{n_k}\to l$ which implies, by using the previous theorem, that $b_{n_k}\to l$, and then $l$ belongs to the adherence values of $b_n$ too. Thus the set of adherence values of $a_n$ is $\subset$ in the set of adherence values of $b_n$, and since $\limsup$ is the maximum value of adherence values and $\liminf$ is the minimum value of adherence values, it is
$$\limsup_{n\to +\infty} a_n\leq \limsup_{n\to +\infty} b_n,$$
which is exactly the converse of what I wanted to prove. Could anyone please help me to understand if I am missing something?
Also an alternative idea is well accepted. Thank you in advance!
