I have read that the notions of convergence in almost and statistical senses are not comparable.

For this both of below must be satisfied :

  1. There is a real convergent sequence statistically, which becomes almost non-convergent.

  2. There is a almost convergent sequence, which becomes statistically non-convergent.

No. 2 is clear to me. Because the sequence (1,0,1,0,1,0,...) is almost convergent to $\frac{1}{2}$. But not statistically convergent. But I am still searching about the validity no.1 statement.

My Qn. : Is there any statistically convergent real sequence, which is not almost convergent?


For the sake of simplicity, let us consider only sequences of zeroes and ones.

Notice that if a sequence contains arbitrarily long segments consisting of consecutive ones and arbitrarily long segments of consecutive zeroes, then it is not almost convergent. This can be seen from the fact that for any fixed $p$ you get both ones and zeroes as the value of the fraction $$\frac{x_{n+1}+x_{n+2}+\dots+x_{n+p}}p.$$

Now it suffices to find a set $A$ such that it has zero density, it contains arbitrarily large segments of consecutive numbers. (The assumption that $A$ has zero density means also that there are arbitrarily large gaps.)

If you choose $$x_n= \begin{cases} 1 & n\in A, \\ 0 & n\notin A, \end{cases} $$ then this sequence is statistically convergent to zero, but it is not almost convergent.

One possible way of looking at this is the following. If a set $A$ determines a sequence $(x_n)$ in the way described above, then it converges statistically to zero if and only if $d(A)=0$, i.e., the asymptotic density is zero.

Such sequence is almost convergent to $L$ if and only if $u(A)=L$, where $u(A)$ denotes Banach density (uniform density). I have collected some references about Banach density and it's relation to asymptotic density in this answer: Density of a set of natural numbers whose differences are not bounded.

So you can look at this problem as searching for a set such that $d(A)=0$ but it does not have Banach density.

Basically the same examples are suggested in the paper H. Miller, C. Orhan: On almost convergent and statistically convergent subsequences; DOI: 10.1023/A:1013877718406, Zbl: 0989.40002, MR1924673

I will reproduce here the relevant part.

Proposition 1.1. Almost convergent and statistical convergence are incompatible; i.e., $\mathbf F\nsubseteqq \mathbf S$ and $\mathbf S\nsubseteq\mathbf F$.

Proof. The sequence $s=(s_n)$ defined by $s_n=1$ if $n$ is even and $s_n=0$ if $n$ is odd is almost convergent to $1/2$, but it is not statistically convergent. Now consider the sequence of $0$'s and $1$'s defined as follows $$\underset{\rightarrow100\leftarrow}{0,\dots,0}\underset{\rightarrow10\leftarrow}{1,\dots,1},0,\dots,0,1,\dots,1$$ where the blocks of $0$'s are increasing by factors of $100$ and blocks of $1$'s are increasing by factors of $10$. This sequence is not almost convergent but is statistically convergent to zero, which completes the proof.

  • $\begingroup$ Do you mean that the sequence $(x_n)_n$ defined by $x_n=1$ if $n$ is a perfect square and $0$ otherwise, is not almost convergent? $\endgroup$ – BijanDatta Jul 3 '19 at 7:44
  • 1
    $\begingroup$ @BijanDatta If you want to use the approach suggested in my answer, you need $A$ which has zero density, but also has arbitrarily long segments of consecutive elements. The set $A=\{n^2; n\in\mathbb N\}$ has zero density, but there aren't any consecutive integers which belong to $A$. $\endgroup$ – Martin Sleziak Jul 3 '19 at 7:47
  • $\begingroup$ Secondly, I've already read this paper of Miller and Orhan. But I could not understand the definition of the second sequence in Proposition1.1 : specially the clause "the blocks of 0's are increasing by factors of 100". I think "multiplier" is appropriate instead of "factor". Is it not? I need your help to understand that sequence. $\endgroup$ – BijanDatta Jul 3 '19 at 7:53
  • 1
    $\begingroup$ @BijanDatta I think that they simply mean that the $k$-th block of zeroes has length $100^k$, the $k$-th block of ones has length $10^k$. Of course, it's possible to choose many other examples. The set $\bigcup_{n=1}^\infty \{n!, n!+1, \dots, n!+n\}$ was suggested on MO. $\endgroup$ – Martin Sleziak Jul 3 '19 at 7:54
  • $\begingroup$ Ok. That means integral power of 100 and 10. $\endgroup$ – BijanDatta Jul 3 '19 at 8:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.