Assuming a number is rational, then proving that this leads to a contradiction, is a powerful tool. I want to know if it can applied in the following case.
Assume a number $n$ is rational, with (fully reduced) form $\frac{a}{b}=\frac{even}{odd}$ for positive integers $\{a,b\}$.
Next assume another fully reduced fraction $\frac{c}{d}=\frac{even}{odd}$, and define $n$ as
$$n=\frac{a}{b}=\lim_{k\to\infty}\frac{c}{d}.\frac{2^kk^{2k}}{(1+k^4)^k}$$ $$=\lim_{k\to\infty}\frac{even}{odd}.\frac{2^kk^{2k}}{(1+k^4)^k}$$
where $k$ takes positive integer values.
For positive integer $k$, it is obvious that $2^kk^{2k}$ is always even. However, $(1+k^4)^k$ is either even or odd depending on the oddness or evenness of $k$: for odd $k$, $(1+k^4)^k$ is even, and vice versa.
Given that we are looking at $\lim_{k\to\infty}$, evaluating $k$ for oddness or evenness becomes meaningless. Yet, deciding whether $\frac{2^kk^{2k}}{(1+k^4)^k}$ is $\frac{even}{odd}$ or $\frac{odd}{even}$ is vital to establishing whether or not integers $a$ and $b$ exist, thereby producing a rational value for $n$.
My question is, does this constitute a proof that there is no rational $\frac{a}{b}$? Or is it merely proof that the question is undecidable from these axioms?
EDIT:
As @Michael and @Saucy O'Path have commented below, $\lim_{k\to\infty}\frac{2^kk^{2k}}{(1+k^4)^k}=0$, which of course I should have spotted. However, my question remains valid, because now we have
$$n=\frac{a}{b}=\lim_{k\to\infty}\frac{c}{d}.\frac{2^kk^{2k}}{(1+k^4)^k}=0$$
But we have defined $n$ thus:
Assume a number $n$ is rational, with (fully reduced) form $\frac{a}{b}=\frac{even}{odd}$ for positive integers $\{a,b\}$.
The assumption demands that $n\ne 0$. But we have just proved that $n=0$. Does this contradiction constitute proof that $n\ne\frac{a}{b}$ and is therefore not a rational number?