Is the sequence $x$ statistically convergent to $0$?

Question

A sequence $y=(y_n)_n$ is said to be a statistically convergent to $\lambda$ if for any $\epsilon>0$ the set $\{n\in\mathbb N: |y_n-\lambda|\geq\epsilon\}$ has natural density $0$.

Consider the following sequence :

$$x=(~\underbrace{0,0,\dots,0}_{100\text{ copies}}~,~ \overbrace{1,1,\dots,1}^{10\text{ copies}}~,~ \underbrace{0,0,\dots,0}_{100^2\text{ copies}}~,~ \overbrace{1,1,\dots,1}^{10^2\text{ copies}}~,~ \underbrace{0,0,\dots,0}_{100^3\text{ copies}}~,~ \overbrace{1,1,\dots,1}^{10^3\text{ copies}}~,~ \dots)$$

This sequence is not almost convergent because there exists no $\Omega\in \mathbb R$ such that $\lim\limits_{k\to \infty}\frac{x_{n+1}+x_{n+2}+\dots+x_{n+k}}{k}=\Omega$, for each $n\in \mathbb N$.

My Question : Is the sequence $x$ statistically convergent to $0$?

How can I prove or disprove that? How we deal here with natural density?

This sequence is from Proposition 1.1 in Miller, H. I.; Orhan, C., On almost convergent and statistically convergent subsequences, Acta Math. Hung. 93, No. 1-2, 135-151 (2001). ZBL0989.40002, MR1924673.

Answer 1

You have a sequence consisting of zeroes and ones. So it is statistically convergent to zero, if and only if the set $$A=\{n\in\mathbb N; x_n=1\}$$ has density zero, i.e., that $d(A)=0$.

Let $A(n)=|A\cap\{1,2,\dots,n\}$| denote the number of elements in $A$ which are less than or equal to $n$. Recall that the asymptotic density is defined as the limit of the fraction $A(n)/n$, taking limes superior/inferior we get the upper/lower asymptotic density.

I will also remind that in general when we have a set which has this type of "block structure", it suffices to check the value of $A(n)/n$ at the beginning end ends of the blocks to get the asymptotic density. See, for example, Proposition 2 in this answer.

In our case, the values of $A(n)/n$ at the ends of the blocks are $$\frac{10}{10+100},\frac{10+10^2}{10+100+10^2+100^2}, \frac{10+10^2+10^3}{10+100+10^2+100^2+10^3+100^3}.$$ Using Stolz-Cesaro theorem we can see that limit of the above sequence is equal to $$\lim_{n\to\infty} \frac{10^n}{10^n+100^n} = \lim_{n\to\infty} \frac{1}{1+10^n}=0.$$ This show that $\overline d(A)=0$ and, consequently, $d(A)=0$.

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¹I am using Stolz-Cesaro theorem in the "sum version", for comparing the two versions of this result see this answer. Of course, it would be easy to calculate the limit even without this theorem - we basically just need to compute the sums of two geometric series.