limsup of average smaller than limsup I have read this solution, but I could not understand it.
It has shown $$\sigma_n\leqslant  \frac 1n\sum_{j=1}^ks_j+\sup_{l\geqslant k}s_l,$$ but how does it go to $$\sup(\sigma_n)\leqslant \frac 1n\sum_{j=1}^ks_j+\sup_{l\geqslant k}s_l$$
I have thought of regarding the RHS as upper bound of $\sigma_n$, but I am not sure about this, since RHS varies as $n$ changes.
The writer wrote "take on both sides the limsup when $n \to \infty$", but I could not understand it. Please help.
 A: First I will copy the relevant part of the answer you are talking about so that we have some context:

Fix an integer $k$. Let $n\geqslant k$. Then 
  $$\sigma_n=\frac 1n\sum_{j=1}^ks_j+\frac 1n\sum_{j=k+1}^ns_j\leqslant \frac 1n\sum_{j=1}^ks_j+\sup_{l\geqslant k}s_l.$$
  Now take on both sides the limsup when $\color{red}{n\to +\infty}$: we get the wanted result.


I would slightly change the formulation and the conclusion. (I think we need to add one more step to finish the conclusion. But I might have missed something obvious. Or maybe Davide considered the missing step easy and left it for the reader.)
So we have the inequality
$$\sigma_n \le \frac 1n\sum_{j=1}^ks_j+\sup_{l\ge k}s_l.$$
This equality is true for any $n\ge k$.
If we take limit superior w.r.t. $n$ on both sides we get
$$\limsup_{n\to\infty} \sigma_n \le \sup_{l\ge k}s_l.$$
(Since the first term on the RHS converges to zero for $n\to\infty$ and the second one does not depend on $n$.)
Now the above inequality is true for arbitrary $k$. So we also have
$$\limsup_{n\to\infty} \sigma_n \le \lim_{k\to\infty} \sup_{l\ge k}s_l = \limsup_{k\to\infty} s_k.$$
A: Here's a simple solution:
Let $x^{*}_k = \sup_{k \geq n} \{x_n\}$. Then, $$x^{*}_k \to \limsup_{n \to \infty} \{x_n\}.$$
By a simple fact, a convergent sequence's averages converge to the same limit, so $$\sigma_n^{*} = \frac{1}{n} \sum_{j=1}^{n} x^{*}_j \to \limsup_{n \to \infty} \{x_n\}$$ as well. Now since $\sigma_n \leq \sigma^{*}_n$, the result follows. 
