Intervals of Limits of Cesàro Means 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?
 A: 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).
