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Is there a bounded real-valued sequence with divergent Cesaro means (i.e. not Cesaro summable)?

More specifically, is there a bounded sequence $\{w_k\}\in l^\infty$ such that $$\lim_{M\rightarrow\infty} \frac{\sum_{k=1}^M w_k}{M}$$ does not exist?

I encountered this problem while studying the limit of average payoffs criterion for ranking payoff sequences in infinitely repeated games.

Martin Sleziak
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tvk
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2 Answers2

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

Feng
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Hanul Jeon
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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.

Paolo Leonetti
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  • Interesting example. Could you tell where you got the motivation for this, or where have you seen something similar maybe? – Anshuman Agrawal Mar 14 '23 at 08:25
  • I think you made a typo in the last to 2nd inequality. I think it should be $\frac{x_1+\cdots+x_{a_1+\cdots+a_{2n-1}}}{a_1+\cdots+a_{2n-1}}\le \frac{a_1+\cdots+a_{2n-2}}{a_1+\cdots+a_{2n-1}}$. – 19021605 Oct 31 '24 at 09:07
  • @AnshumanAgrawal I think this may be the motivation behind the construction. Once the quotient is less than $\frac{1}{3}$, add $1$'s to the numerator until the quotient is greater than $\frac{2}{3}$. Once the quotient is greater than $\frac{2}{3}$, don't add $1$'s to the numerator (i.e. adding $0$'s) while allowing the denominator to grow until the quotient becomes less than $\frac{1}{3}$. – 19021605 Oct 31 '24 at 09:13