# Does the convergent Cesàro mean of a sequence imply that the sequence is bounded?

It is well known that a convergent sequence implies that its Cesàro mean converges while the converse is wrong. Then I am wondering the following:

1. Would a bounded sequence imply the convergence of its Cesàro mean?
2. Would the convergent Cesàro mean of a sequence imply that the sequence is bounded?

It would be appreciated if any proofs and examples are provided.

1. Consider the sequence $1,-1,1,1,-1,-1,1,1,1,1,-1,-1,-1,-1,1,1,1,1,1,1,1,1,-1,-1,-1,-1,-1,-1,-1,-1,\ldots$ where we have runs of $1$s or $-1$s with length $2^k$. It is a bounded sequence, but its Cesaro means are not convergent;
2. Consider the sequence $\{a_n\}_{n\geq 1}$ where $a_n=1$ unless $n=2^{m}$ for some $m\in\mathbb{N}$, where $a_n=\log_2(n)=m$. The Cesaro means of $\{a_n\}_{n\geq 1}$ converge to $1$, but $\{a_n\}_{n\geq 1}$ is an unbounded sequence.
1. No. Let $\{a_n\}$ be defined as follows: $a_1=1$. $a_2=a_3=...=a_{n_1}=-1$, where $n_1>1$ is such that $|\frac{a_1+a_1+...+a_{n_1}}{n_1}-(-1)|<1/2$. Define $a_{n_1+1}=a_{n_1+2}=...=a_{n_2}=1$ where $a_{n_2}$ is such that $|\frac{a_1+a_1+...+a_{n_2}}{n_2}-1|<1/2$. And so on, inductively.
2. No. Let $a_n=\sqrt{n}$ when $n$ is a perfect square and $0$ otherwise. It is easy to see that $\frac{a_1+...+a_n}{n}\to0$