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.
Best Answer
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.