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?

let $L = \lim \sup s$
if $L = \infty$ there is nothing to prove. if $L$ is finite there is no loss of generality in assuming $L=0$
given $\epsilon \gt 0$ then $\exists N. n \gt N \Rightarrow s_n \lt \frac12 \epsilon$
$$
\sigma_n \lt \frac1{n}\left(\sum_{k=1}^N s_k +\frac12(n-N)\epsilon \right)
$$
set $M$ so that $\frac1{M} \sum_{k=1}^Ns_k \lt \frac12 \epsilon$
then for $n \gt M$ we have
$$
\sigma_n \lt \epsilon
$$
hence 
$$
\lim \sup \sigma \lt \epsilon
$$
