2
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.