Proving that $\limsup_{n\to\infty}\frac{1}{n}\sum_{m=1}^n s_m\leq \limsup_{n\to\infty}s_n.$ I am reviewing for my first year analysis exam and am stuck on a problem.
Let $\sigma_n=\frac{1}{n}\sum_{m=1}^n s_m$.  I am trying to show that,

if $(s_n)$ is a bounded sequence of real numbers, $$\limsup_{n\to\infty}\sigma_n\leq \limsup_{n\to\infty}s_n.$$

I understand what's supposed to happen here, but I'm having trouble showing it.  This is what I have so far:
I know from a previous exercise that $s_n\rightarrow s$ implies that $\sigma_n\rightarrow s$. I'll call this fact $\star$.
If $(\sigma_{n_k})_{k\in K}$ is a convergent subsequence of $(\sigma_n)$, let $(s_{n_i})_{i\in I\subseteq K}$ be a convergent subsequence of $(s_{n_k})_{k\in K}$.  We know this must exist because $(s_{n_k})_{k\in K}$ is bounded (because $(s_n)$ is bounded).  Let $I_k=\{i\in I:i \leq k\}$, $I_k^\prime =\{i\in K\setminus I:i \leq k\}$, and $m_k=|I_k|$.  Then $$\sigma_{n_k}=\frac{m_k}{n_k}\left(\frac{1}{m_k}\sum_{i\in I_k}s_{n_i}\right)+\frac{n_k-m_k}{n_k}\left(\frac{1}{n_k-m_k}\sum_{i\in I_k^\prime}s_{n_i}\right)$$
Because $s_{n_i}$ is convergent by assumption, $\frac{1}{m_k}\sum_{i\in I_k}s_{n_i}$ is convergent by $\star$.
But, now I don't know where to go.  The coefficients are a little ugly and I don't know what the other sum might converge to.  What should I do next?  Or am I taking the wrong approach?

My intuition is that this the wrong approach, and I would rather look at the proof of fact $\star$ for inspiration than the statement of fact $\star$.
What happens if you take a subsequence of $\sigma_n$ that converges to $\limsup\sigma_n$ and try to find indices for an $s_n$-subsequence that converges to something $\ge\limsup\sigma_n$?
