Theorem by Cesàro: how to prove? The following was introduced to me in class as «an old theorem, almost forgotten, by an old mathematician, Cesàro, 1960».
Theorem
Suppose $\{a_n\}$ is a real sequence. Set:
$$A_n=\frac{1}{n+1}\sum_{k=1}^na_k.$$
Then:
$$\liminf_na_n\leq\liminf_NA_N\leq\limsup_NA_N\leq\limsup_na_n.$$
I tried googling for it, but was unable to find it. I cannot seem to be able to prove it myself. How would I go about this?

We'll prove that $$\liminf a_n\leq \liminf A_n$$
If $\liminf a_n=-\infty$ this is obviously true.
Otherwise, let $\alpha<\liminf a_n$. We will show that $\alpha\leq\liminf A_n$.
By the definition of $\liminf$, this means that there exists a $K$ such that for all $n>K$, $\alpha<a_n$. 
Now, if $N>K$ then:
$$A_N = \frac{1}{N+1}\left((K+1)A_K+\sum_{n=K+1}^{N} a_k\right)>\frac{1}{N+1}\left((K+1)A_K+(N-K)\alpha\right)$$
Now, $\lim_{N\to\infty} \frac{(K+1)A_K}{N+1} =0$. And $\lim_{N\to\infty} \frac{N-K}{N+1}=1$. So $\liminf A_n\geq \alpha$.
