# Equivalence on Cesàro convergence

Let $$(x_n)$$ be a bounded real sequence. Then $$\lim_{n\to \infty}\frac{1}{n}\sum_{i=1}^n x_i=0 \Longleftrightarrow \lim_{n\to \infty}\frac{1}{2^n}\sum_{i=2^n+1}^{2^{n+1}} x_i=0.$$ Do you have a reference for this equivalence? I remember that I saw it years ago here on StackExchange, but I couldn't find it

EDIT: The above equivalence does not hold: A possibile counterexample is given by the sequence $$(x_n)$$ defined by $$x_n=1$$ if $$2^k\le n\le \frac{3}{2}2^k$$, and $$x_n=-1$$ otherwise.

My proof: Set $$a_n:=\frac{1}{n}\sum_{i=1}^n x_i$$ and $$b_n:=\frac{1}{2^n}\sum_{i=2^n+1}^{2^{n+1}}x_i$$ for all $$n\ge 1$$.

Suppose that $$a_n\to 0$$; then $$b_n \to 0$$ because $$\frac{a_{2^n}+b_n}{2}=a_{2^{n+1}}$$. Conversely, suppose that $$b_n\to 0$$. Then $$a_n \to 0$$ because, if $$2^k then $$na_n=\sum_{i=1}^nx_i=x_1+x_2+2b_1+4b_2+\cdots+2^{k-1}b_{k-1}+x_{2^k+1}+\cdots+x_n,$$ which implies that $$a_n$$, up to negligible terms, is an average of these $$b_i$$ and "a part of $$b_k$$" which is negligible by itself. (The bolded part is wrong. It works if the sequence is nonnegative.)

• I'm going to ask a separate question to check my proof. – user Oct 31 '20 at 17:54
• @user Well, wait: it couldn't be true as it is. The limit had to be zero.. – Paolo Leonetti Oct 31 '20 at 17:54
• Yes I've changed it acconrdingly. But the way and doubts are the same. – user Oct 31 '20 at 17:55
• The question is here if you ar einterested to aswer it. Thanks – user Oct 31 '20 at 18:00
• Ok, I move the discussion there – Paolo Leonetti Oct 31 '20 at 18:00