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Consider an infinite sequence $(a_k)$. I am interested in the limit of Cesàro means of that sequence—i.e., $$\lim_{n\to\infty}\frac1n\sum_{k=1}^na_k$$ and of sequences that are rearrangements of the terms of $(a_k)$.

For example, the consider the sequences:

$$A = (1, 0, 1, 0, 1, 0, \dots)$$ $$B = (1, 0, 0, 1, 0, 0, \dots)$$

The limit of Cesàro means of $A$ is $\frac{1}{2}$. The limit of Cesàro means of $B$ is $\frac{1}{3}$.

Presumably, for any rearrangement of the sequences above, the limit of Cesàro means (if it exists) will be in $[0,1]$. Can it attain any value in that interval?

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  • $\begingroup$ If the average over all n is 1/2, it shouldn't matter in which order you add them. $\endgroup$ – Dan Christensen Mar 15 '16 at 21:25
  • $\begingroup$ I don't want to add them. I'm only concerned with the means. $\endgroup$ – bakeryjake Mar 16 '16 at 0:22
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By rearrangement, I guess you mean you form a new sequence $b_n$ such that $b_n = a_{\sigma(n)}$ for some bijection $\sigma : ℕ → ℕ$. Note that the two sequences you provided are by this definition, a rearrangement of each other. (The 'paradox' of the increase in the number of zeros is due to the same reason that there are 2 times as many naturals as there are evens.)

By choosing this bijection, we can indeed choose the Cesaro limit to be anything we please in $[0,1]$. To obtain a Cesaro mean of $x∈(0,1)$, include 1's until you reach a partial Cesaro mean larger than $x$, $$\frac{1}{n} \sum_{k=1}^n b_k > x$$
then start setting $b_k$s to be $0$s until $$\frac{1}{n} \sum_{k=1}^n b_k < x$$ This process can be clearly continued ad infinitum, and the Cesaro mean is indeed $x$.

To approach 0, use $1,0,1,0,0,1,0,0,0,0,1,…$ i.e. with exponentially increasing lengths of $0$s. This exponential growth will beat the 'averaging'; we see that if $N$ is such that $2^{N-1} ≤ n < 2^N$,

$$ \frac{1}{n}\sum_{k=1}^n a_k ≤ \frac{N-1}{n} ≤ \frac{\log_2 n}{n} → 0$$

One deals with $1$ similarly.


A comment has asked for clarification.

We start with the sequence $(a_n) = (1,0,1,0,…)$, where every odd element is 1 and every even element is 0. i.e. $a_{2k-1} = 1$ and $a_{2k}=0$ for $k\geq 1$. We repeatedly exploit the fact that if a sequence $c_n$ is eventually constant, $$c_n = c \quad n\gg 1$$ then its Cesaro mean is $c$.

We now construct the $\sigma$ that will give us a mean $x∈(0,1)$. Set $n^-_1=1$, and define $\sigma(1) = 2$. Then set $$\sigma(n^-_1 + 1)=1,\sigma(n^-_1 + 1) = 3,…,\sigma(n^-_1 + n^+_1) = 2n^+_1 - 1$$ where $n^+_1$ is the smallest natural such that $$ \frac{1}{n^-_1 + n^+_1}\sum_{k=n^-_1 + 1}^{n^+_1} 1 > x$$ Such an $n^+_1$ exists by the fact stated above. Now start setting $\sigma$s to be even, $$ \sigma(n^-_1 + n^+_1 + 1) = 4, … \sigma(n^-_1 + n^+_1 + n^-_2) = 2(n^-_1 + n^-_2)$$ where $n^-_2$ is the smallest natural such that $$ \frac{1}{n^-_1 + n^+_1 + n^-_2}\sum_{k=n^-_1 + 1}^{n^+_1} 1 < x$$ We then start including more odd numbers, and then more even numbers, and so on. By construction, we see that for each $m$, $\sigma(1),\sigma(2),…$ eventually takes all even values less than $$2(n^-_1 + … + n^-_m)$$ and all odd values before $$2(n^+_1 + … + n^+_m)-1$$ (surjective) and it never repeats (injective).

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  • $\begingroup$ Are you sure that the $b_k$ are actually a permutation of the $a_k$ ? $\endgroup$ – Gabriel Romon Mar 14 '16 at 22:44
  • $\begingroup$ @LeGrandDODOM as long as you include infinitely many $0$s and $1$s, yeah they are. $\endgroup$ – Calvin Khor Mar 14 '16 at 22:45
  • $\begingroup$ it doesn't seem obvious to me, could you elaborate ? $\endgroup$ – Gabriel Romon Mar 14 '16 at 22:48
  • $\begingroup$ @LeGrandDODOM I have added some more explicit reasoning, does it help? $\endgroup$ – Calvin Khor Mar 14 '16 at 23:12
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    $\begingroup$ @bakeryjake I doubt you can create a sequence whose rearrangements have an unbounded range of possible averages; the few candidates I thought of are not Cesaro summable e.g. $a_n = n(-1)^n $, and I have also not seen an unbounded yet Cesaro summable sequence. One scenario not covered by the above however is that its possible for the averages to not converge at all; this is possible by making the partial averages sway like a sine wave, so to speak :) $\endgroup$ – Calvin Khor Mar 15 '16 at 0:30

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