I will try to ask my question as clear as possible.

We know that, there exist infinitely number of infinite sequences that, consist of elements $\left\{0,1,2 \right\}$, which is can not express by the any closed-form expression or any specific mathematical function.

I don't know definition of such a sequences. I know only, such sequences are exist.

Let, $A_n=\left\{a_1,a_2,a_3,\cdots, a_{n\to\infty}\right\}$ sequence be an infinite sequence, where $i≥1, ∀ a_i\in\left\{0,1,2\right\}$.

I define this infinite sequence as an sequence such that selected from an uncountable infinite set that cannot be given by any mathematical function.

This is obvious,

$$0≤\lim_{n\to\infty} \frac{\sum_{i=1}^{n}a_n}{2n}≤1$$

Because, $∀ a_i≤2.$

If, we choose the infinite sequence $a_n$, from a countable infinite set, we can write (for example),

$$a_n=n+2-3 \left \lfloor {\frac{n+2}{3}}\right \rfloor $$

Then, for $\sum_{i=1}^{n}a_n$ I have an exact form:

$$\sum_{i=1}^{n}a_n=\left \lfloor{\frac{n - 2}{3}}\right\rfloor + 2 \left(\left\lfloor{\frac n3}\right\rfloor + 1 \right) + 1$$

Therefore, we have

$$\lim_{n\to\infty}\frac{\sum_{i=1}^{n}a_n}{2n}=\frac{\left \lfloor{\frac{n - 2}{3}}\right\rfloor + 2 \left(\left\lfloor{\frac n3}\right\rfloor + 1 \right) + 1}{2n}=\frac 12$$

It is obvious, if, we choose the infinite sequence $a_n$, from a uncountable infinite set, this is impossible to write an exact form for

$$\lim_{n\to\infty} \frac{\sum_{i=1}^{n}a_n}{2n}$$

We have only

$$0≤\lim_{n\to\infty} \frac{\sum_{i=1}^{n}a_n}{2n}≤1$$

Finally, I want to ask my question:

For any arbitary constant $0≤\alpha≤1$, can we say that there exist such an infinite sequence, which is selected from an uncountable set and not expressed by any mathematical function, such that

$$\lim_{n\to\infty} \frac{\sum_{i=1}^{n}a_n}{2n}=\alpha \text{(in exact form)}$$

But, the sequence $A_n$,can not given by an any exact function.

Is this statement correct?

The meat of question, I'm trying to say,

There exist such a infinite sequence $A_n$,

$$\lim_{n\to\infty} \frac{\sum_{i=1}^{n}a_n}{2n}=\text{exact form constant} $$

But, we can never define the sequence $A_n$ as a sequence given by any mathematical function/ $n-$th term closed form/ recurrence formula/ algorithm and etc.

Is this claim correct?

Thank you.

  • $\begingroup$ One thing you could try, is prove that $$\left\{\lim_{n\to\infty}\frac{\sum_{i=1}^na_n}{2n} \mid a_i\in\{0,1,2\}\right\}=[0,1]$$ which would prove your claim, if I understand you correctly $\endgroup$ – vrugtehagel Jun 10 '19 at 10:41
  • $\begingroup$ @vrugtehagel Do you think my question is understandable? $\endgroup$ – lone student Jun 10 '19 at 10:45
  • $\begingroup$ Is the following what you are asking? Let $a_1, a_2, \ldots$ be a sequence whose terms are taken from the set $\{0,1,2\}$. Let $0\leq\alpha\leq 1$. Do there exist infinitely many such sequences, such that $\lim_{n\rightarrow\infty}\frac{\sum_{i=1}^na_i}{2n}=\alpha$, but such that $\frac{\sum_{i=1}^na_i}{2n}$ has no closed form expression? $\endgroup$ – Mankind Jun 10 '19 at 10:50
  • $\begingroup$ @Mankind I mean, there exist such infinite sequence, for $\alpha$ we have an exact form, but for $A_n$, we don't have. $\endgroup$ – lone student Jun 10 '19 at 11:34
  • 1
    $\begingroup$ Downvote without comment..?please specify points that are not clear in the comments. $\endgroup$ – lone student Jun 10 '19 at 17:03

I shall assume that with "cannot be given by a mathematical function" you mean that there does not exist a finite length formula (or algorithm) that describes the function exactly.

First, there exists such a sequence where we even have $a_i\in\{0,2\}$ for each $i$: Just ensure that $\sum_{i=1}^n a_i=2\lfloor \alpha n\rfloor$ by letting $$a_n=\begin{cases}0&\lfloor \alpha n\rfloor =\lfloor \alpha (n-1)\rfloor\\ 2 & \text{otherwise} \end{cases} $$

Unfortunately, if $\alpha$ is computable, this $a_n$ is "given by a mathematical function" - not what you want.

If $0<\alpha<1$, we will have infinitely many $n$ with $a_n=0$ and infinitely many $n$ with $a_n=2$, hence also inifinitely many $n$ with $a_n=0$ and $a_{n+1}=2$. It is allowed to replace an arbitrary subset of these cases these with $a_n=a_{n+1}=1$ without changing the limit behaviour. As there are uncountably many such subsets, not all can be described the only countably many formulas, hence at least some of these "cannot be given by a mathematical function".

We are still left with the special cases $\alpha=0$ and $\alpha=1$. For $\alpha=1$, we would start with all $_n=0$ and so lack the infinitely many $0$-$2$ steps we used to produce uncountably many sequence variations. However, if we set $a_n=2$ whenever $n$ is a perfect square, we still have limit $0$ but now have infinitely many cases of $a_n=0$, $a_{n+1}=2$ again, and can continue as above. The same argument works for $\alpha=1$ with the roles of $0$ and $2$ exchanged.


Yes, every $\alpha$ can be expressed. Note that appending a new $2$ at the end of the sequence must increase the average, and appending a new $0$ at the end must decrease the average. So if our sum is below $\alpha$, we simply add a $2$ at the end, and if it is above we simply add $0$. Note that the total change to the sum approaches $0$, so our sequence must converge, but also note that we can definitely reach above $\alpha$ starting from below since adding a string of $2$ yields $1$, and similarly we can reach below $\alpha$ starting from above. So we can approach $\alpha$ with arbitrary precision.

  • $\begingroup$ But, $A_n$, can not express by any Mathematical function. $\endgroup$ – lone student Jun 10 '19 at 10:50

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.