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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?

bakeryjake
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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).

Calvin Khor
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  • Are you sure that the $b_k$ are actually a permutation of the $a_k$ ? – Gabriel Romon Mar 14 '16 at 22:44
  • @LeGrandDODOM as long as you include infinitely many $0$s and $1$s, yeah they are. – Calvin Khor Mar 14 '16 at 22:45
  • it doesn't seem obvious to me, could you elaborate ? – Gabriel Romon Mar 14 '16 at 22:48
  • @LeGrandDODOM I have added some more explicit reasoning, does it help? – Calvin Khor Mar 14 '16 at 23:12
  • @LeGrandDODOM: For an explicit permutation, assume $k=0,1,2,\ldots$ and write $$b(k) = \begin{cases} a(2k/3) & \text{if }k \equiv 0\ (\text{mod }3) \ a(4(k-1)/3 + 1) & \text{if }k \equiv 1\ (\text{mod }3) \ a(4(k-2)/3 + 3) & \text{if }k \equiv 2\ (\text{mod }3) \ \end{cases}$$ –  Mar 14 '16 at 23:50
  • Thank you, @LeGrandDODOM. I'm interested in this same question for other sequences, too. Are there sequences where the averages can converge to different values for different rearrangements (in the sense you've defined), without them always converging to any value within some closed interval? (I'm sorry if this question makes no sense.) – bakeryjake Mar 15 '16 at 00:22
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    @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 :) – Calvin Khor Mar 15 '16 at 00:30