# Limsup inequality (Arithmetic Mean)

Suppose $\{a_n\}$ is a bounded sequence of real numbers. How do we prove that $$\limsup_{n\to\infty}\frac{a_1+a_3+\dots+a_{n+1}}{n}\leq\limsup_{n\to\infty}a_n$$?

Note that there is a "skipped" term $a_2$ in the sum above.

I am not experienced in proving inequalities involved lim sup, what I can observe is that $$\frac{a_1+a_3+\dots+a_{n+1}}{n}\leq\max\{a_1, a_3, \dots, a_{n+1}\}.$$

I also thought of showing $$\sup_{n\geq j}\frac{a_1+a_3+\dots+a_{n+1}}{n}\leq\sup_{n\geq j}a_n$$, then take limits as $j\to\infty$.

Thanks for any help!

Look at proofs for the theorem of Cesaro-Stolz.

That you omit $a_2$ does not play any role, just consider the modified sequence were $\tilde a_a=a_1$, $\tilde a_k=a_{k+1}$.

For any $ε>0$ there is an $N$ such that for any $n\ge N:$ $a_n<\limsup_ka_k+ε$. Now split the sum on the left at $N$, etc.

Let $L=\limsup_{n\to\infty}a_n<\infty$. Let $\epsilon>0$. By definition of $\limsup$, there exists $K$ such that $a_n<L+\epsilon$ for all $n>K$.

Then $$\frac{a_1+a_3+\dots+a_{n+1}}{n}<\frac{a_1+a_3+\dots+a_K+(L+\epsilon)(n-K+1)}{n}.$$

Taking $\limsup_{n\to\infty}$ on both sides gives $$\limsup_{n\to\infty}\frac{a_1+a_3+\dots+a_{n+1}}{n}\leq L+\epsilon.$$

Since $\epsilon>0$ is arbitrary, that proves the statement.