Is it possible to find such a function $f:\mathbb{N_{>0}}\longrightarrow \left\{0,1,2 \right\}$ or a sequence $f(n):= x_n$ and such a function $g:\mathbb{N_{>0}}\longrightarrow \left\{0,1,2 \right\}$ or a sequence $g(n):= y_n$ which gives
$$\lim_{n \to \infty }\text{sup}\dfrac{\displaystyle \sum_{k=1}^{n} \left(5^k \times f(k) \right)}{\displaystyle \sum_{k=1}^{n} \left(5^k \times g(k) \right)}=\infty$$
I know only trivial solutions, for example let, $g(k)$ be a constant function, where $g(k)=0$ and let $f(k)$ be a constant function, where $f(k)=2$.
But I`m looking for a non-constant functions which provide the conditions of the problem. How can I find such non-trivial functions?
I tried $f(k)= k\mod3$ $\quad$ and $\quad$ $g(k)=k^2 \mod3 $
$f(k)= k^2 \mod 2$ $\quad$ and $\quad$$g(k)=k \mod 3$ $\thinspace$ and similar elementary functions, which that doesn`t work.
Suppose $f(k)\ne 0$ when $k$ is $1$ less than a perfect square, but $g(k)=0$ except when $k$ is a perfect square. When $n=(m+1)^2-1$ we have $$\sum_{k=1}^n5^kf(k)\ge 5^nf(n)\ge 5^n=5^{m^2+2m}\quad .... (I)$$ but $$\sum_{k=1}^n5^kg(k)= \sum_{j=1}^m5^{j^2}g(j^2)\le m\cdot 5^{m^2}\cdot 2\quad ....(II).$$ The ratio of the far RHS of $(I)$ to $(II)$ is $\frac {5^{2m}}{2m},$ which $\to \infty$ as $m$ (... and hence also as $n=(m+1)^2-1 ...)\to \infty.$