# Limit superior and inferior of Cesàro means

From a sequence $a^x_n$, define the sequence of its Cesàro means $a^{x+1}_n$ as $$\sum_{k=1}^{n} a^x_k/n$$ It is easy to show that the sequence of Cesàro means of a bounded sequence will itself be bounded. It is also easy to construct bounded sequences such that the limit superior and limit inferior of the sequence of their Cesàro means are the same as the limit superior and limit inferior of the original sequence. For example, start with 0, add 1's until you reach a Cesàro mean of 1/2, add 0's until you reach 1/3, add 1's until you reach 3/4, add 0's until you reach 1/5, etc. Now, is it always the case that: $$\limsup_{n\to\infty} a^x_n=\limsup_{n\to\infty} a^{x+1}_n\to\limsup_{n\to\infty} a^{x+1}_n=\limsup_{n\to\infty} a^{x+2}_n$$ if not, is it always the case that: $$\limsup_{n\to\infty} a^x_n=\limsup_{n\to\infty} a^{x+1}_n\land\liminf_{n\to\infty} a^x_n=\liminf_{n\to\infty} a^{x+1}_n$$ $$\to\limsup_{n\to\infty} a^{x+1}_n=\limsup_{n\to\infty} a^{x+2}_n\land\liminf_{n\to\infty} a^{x+1}_n=\liminf_{n\to\infty} a^{x+2}_n$$

• Your second sentence is incomprehensible to me. – zhw. Aug 24 '17 at 19:32
• Please consider adding relevant definitions to your question. Like for the "Cersàro means of a bounded sequence", I assume you mean $\frac{1}{N}\sum^{N}_{i=1}a_i$ for each $N\in\mathbb{N}$ where $(a_n)$ is some bounded sequence in $\mathbb{R}$ ? – Walt van Amstel Aug 26 '17 at 9:51