Cesaro limits and supremum of a sequence

Question

$a_n$ be a sequence of real numbers with finite Cesaro limit, i.e., $\frac{a_1+ \cdots + a_n}{n} \rightarrow c$ for some real number $c$. If $M_n= \max\{a_1, \cdots, a_n\}$, then does $\frac{M_n}{n} \rightarrow 0$?

I believe the answer is false and you can come up with alternating series with increasing absolute value such that the supremum may occur at last term and get non-zero limit. But, I’m unable to cook up such example. Thanks in advance.

Answer 1

The given condition implies that

$$ \frac{a_n}{n} = \frac{a_1 + \cdots + a_n}{n} - \frac{n-1}{n}\cdot\frac{a_1 + \cdots + a_{n-1}}{n-1} \xrightarrow[n\to\infty]{} c - c = 0. $$

Now we can appeal to the following fact to show that $\lim_{n\to\infty} M_n/n = 0$.

Proposition. Let $(b_n)$ be an increasing sequence of positive reals such that $b_n \uparrow \infty$. Then for any sequence of real numbers $(a_n)$, we have

$$ \limsup_{n\to\infty} \frac{\max\{a_1,\cdots,a_n\}}{b_n} = \max\left\{ 0, \limsup_{n\to\infty} \frac{a_n}{b_n} \right\}. $$

Proof. We show that one side bounds the other. Since $\geq$ is clear, we only prove the other direction. Write $M_n = \max\{a_1, \cdots, a_n\}$.

• If $(a_n)$ is bounded above, then clearly $M_n/b_n \to 0$ and $\limsup a_n/b_n \leq 0$.

• Otherwise, for each $k$, define $n_k$ as the largest $n \in \{1,\cdots,k\}$ for which $a_n = M_k$ holds. Since $(a_{n_k})$ is eventually positive, for large $k$ we have

  $$ \frac{M_k}{b_k} = \frac{a_{n_k}}{b_k} \leq \frac{a_{n_k}}{b_{n_k}}. $$

  Letting $k\to\infty$ yields $\limsup_{k\to\infty} M_k/b_k \leq \limsup_{n\to\infty} a_n/b_n$.