Is it possible to find such a computable function $f(n)$, there doesn't exist a limit point for $\lim_{n\to\infty} \frac{\sum_{k=1}^{n}f(k)}{2n} ?$

Question

Let, $a_n=\left\{a_1,a_2,a_3,\cdots, a_{n\to\infty}\right\}$ be a computable infinite sequence, or $f(n)=\left\{a_n\right\}_{n\in\mathbb N}$ be a computable function, where $i≥1, ∀ a_i\in\left\{0,1,2\right\}$.

We have $$0≤\lim_{n\to\infty} \frac{\sum_{k=1}^{n}a_k}{2n}≤1$$

For example:

Let, $$a_n=\left\{0,1,2,0,1,2,0,1,2\cdots \cdots\cdots\right\}$$ or $$f(n)=n+2-3 \left \lfloor {\frac{n+2}{3}}\right \rfloor$$

Then, we get

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

So, we have

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

This result imply, if $f(n)=n+2-3 \left \lfloor {\frac{n+2}{3}}\right \rfloor$ there exist a stationary limit point for $\lim_{n\to\infty} \frac{\sum_{k=1}^{n}f(k)}{2n}$, which equals to $\frac 12.$

And here is my question:

If the function $f(k)$ is computable, is there always a limit point for $\lim_{n\to\infty} \frac{\sum_{k=1}^{n}f(k)}{2n}$ or is it possible to find such a computable function $f(n)$, there doesn't exist a limit point for $\lim_{n\to\infty} \frac{\sum_{k=1}^{n}f(k)}{2n}$ ?

I mean, $$\lim_{n\to\infty}\text{sup} \frac{\sum_{k=1}^{n}f(k)}{2n}≠\lim_{n\to\infty} \text{inf}\frac{\sum_{k=1}^{n}f(k)}{2n}$$

Answer 1

Here is mine. $f(k) = 0$ if $k$ has an even number of digits, and $f(k) = 1$ if $k$ has an odd number of digits.

So:
$f(k) = 1$ for $k$ from $1$ to $9$, since $k$ has $1$ digit and $1$ is odd.
$f(k) = 0$ for $k$ from $10$ to $99$, since $k$ has $2$ digits and $2$ is even.
$f(k) = 1$ for $k$ from $100$ to $999$, since $k$ has $3$ digits and $3$ is odd.
$f(k) = 0$ for $k$ from $1000$ to $9999$, since $k$ has $4$ digits and $4$ is even.

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Now let's do some computations for special values of $n$. First consider $n = 10^z-1$ with $z$ even. Then
$$ f(k) = 0, \quad 10^{z-1} \le k < 10^z, \\ f(k) \le 1, \quad 1 \le k < 10^{z-1} $$ so $$ \sum_{k=1}^n f(k) \le 10^{z-1} = \frac{n}{10} \\ \frac{1}{2n}\sum_{k=1}^n f(k) \le \frac{1}{20} . $$ From this we conclude $$ \liminf_{n \to \infty}\frac{1}{2n}\sum_{k=1}^n f(k) \le \frac{1}{20} $$

Next consider $n = 10^z-1$ with $z$ odd. Then
$$ f(k) = 1, \quad 10^{z-1} \le k < 10^z, \\ f(k) \ge 0, \quad 1 \le k < 10^{z-1} $$ so $$ \sum_{k=1}^n f(k) \ge 10^z - 10^{z-1} = \frac{9 n}{10} \\ \frac{1}{2n}\sum_{k=1}^n f(k) \ge \frac{9}{20} . $$ From this we conclude $$ \limsup_{n \to \infty}\frac{1}{2n}\sum_{k=1}^n f(k) \ge \frac{9}{20} $$ Finally, $$ \lim_{n \to \infty}\frac{1}{2n}\sum_{k=1}^n f(k)\qquad\text{does not exist.} $$

Answer 2

Define

$$a_{n+1}=\begin{cases}0,&\frac1n\sum_{k\le n}a_k>\frac32\lor\left(\frac1n\sum_{k\le n}a_k>\frac12\land a_n=0\right)\\2,&\text{else}\end{cases}$$

This gives us a computable sequence of $0$'s and $2$'s that have both $\frac14$ and $\frac34$ as limit points, or more simply, we have $a_n<\frac14$ for infinitely many $n$ and $a_n>\frac34$ for infinitely many $n$.