Bounded sequence with divergent Cesaro means

Consider $1,-1,-1,1,1,1,1,-1,-1,-1,-1,-1,-1,-1,-1,\cdots$ (one $1$, two $-1$, four $1$, eight $-1$, ...) Then $$\frac{1-2+2^2-2^3+\cdots+(-2)^n}{1+2+2^2+\cdots+2^n}=\frac{(-2)^{n+1} -1}{3(2^{n+1}-1)}$$ This sequence is divergent. So $(\sum_{k\le M}a_k)/M$ has divergent subsequence, and it implies nonexistence of Cesaro mean of $a_n$.

Let $(a_n)$ be any increasing sequence of positive integers such that $$ a_{n+1}\ge 2(a_1+\cdots+a_{n}) $$ for all $n$. Now let $(x_n)$ be the sequence $$ \underbrace{0,\ldots,0}_{a_1 \text{ times }},\underbrace{1,\ldots,1}_{a_2 \text{ times }},\underbrace{0,\ldots,0}_{a_3 \text{ times }},\underbrace{1,\ldots,1}_{a_4 \text{ times }},\ldots $$ It easily follows that $$ \frac{x_1+\cdots+x_{a_1+\cdots+a_{2n}}}{a_1+\cdots+a_{2n}}\ge \frac{a_{2n}}{a_1+\cdots+a_{2n}}=1-\frac{a_1+\cdots+a_{2n-1}}{a_1+\cdots+a_{2n}}\ge 1-\frac{1}{1+2}=\frac{2}{3} $$ for all $n$ and, on the other hand, $$ \frac{x_1+\cdots+x_{a_1+\cdots+a_{2n-1}}}{a_1+\cdots+a_{2n-1}}\le \frac{a_1+\cdots+a_{2n-1}}{a_1+\cdots+a_{2n}} \le \frac{1}{1+2}=\frac{1}{3}, $$ hence it cannot converge.