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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 $2^k<n\le 2^{k+1}$ 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.)

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

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