I am confused at the very core my understanding of numbers for this particular subject, for some $n \in \mathbb N \backslash {\{1}\}$,the following appears to become an increasingly accurate approximation for any $x \in \mathbb R$, regardless of it's algebraicity or rationality:
$$\lim _{k\rightarrow \infty}\Biggl(n^{\lfloor\ln_n(x)\rfloor-k+1}\Bigl \lfloor xn^{k-\lfloor\ln_n(x)\rfloor-1}\Bigr\rfloor\Biggr)\approx{\{x}\}$$
Take for example, $n=2$ and $x=\pi^2$, if $k \lt N$ for some $N$,the difference between the expression inside the limit on left hand side and the fractional part of $\pi^2$ on the right hand side approaches $0$ as $N$ is made larger, which, by the epsilon test, would mean the limit is indeed true and we can replace the approximation sign with an equality.
But how can this be true? the left hand side is clearly rational, and ${\{\pi^2}\}$ is of course a transcendental number.
So an example of my assertion is that:
$$\lim _{k\rightarrow \infty}\Biggl(2^{\lfloor\frac{\ln(\pi^2)}{\ln(2)}\rfloor-k+1}\Bigl \lfloor {\pi}^{2}2^{k-\lfloor\frac{\ln(\pi^2)}{\ln(2)}\rfloor-1}\Bigr\rfloor\Biggr)=\pi^2-9$$
And I haven't been able to get my version of maple to make a float approximation, (and I am not allowed to know the specifics of the method for which the software makes float approximations anyway, as it is inbuilt code) the rational value that is computed for increasingly large values of $N$ for:
$$\lim _{k\rightarrow N}\Biggl(2^{\lfloor\frac{\ln(\pi^2)}{\ln(2)}\rfloor-k+1}\Bigl \lfloor {\pi}^{2}2^{k-\lfloor\frac{\ln(\pi^2)}{\ln(2)}\rfloor-1}\Bigr\rfloor\Biggr)$$
Does appear to support the assertion, however rigor is missing in a proof of such.
So my question is, how can I reconcile this contradiction? Is this an error in intuitive judgement I have made?
Later I will add the epsilon definition of a limit and assign the values appropriate to this example to confirm this to be true, and try to construct an epsilon-delta proof that will hopefully lead me to a stronger conclusion. I am not very good with these constructions, so help is appreciated greatly.
I guess another way Id like to put my question. can I somehow use the expression above as some kind of rationality test based on an epsilon-delta construction as per the definition of a limit?
This particular case is of special interest to me:
As the above expression is always zero if $N \lt \infty$.
Best Answer
No epsilon-delta test can distinguish between rational and irrational numbers. Every interval contains infinitely many of those. It is easy to find infinite sums of rationals that converge to an irrational and infinite sums of irrationals that converge to rationals.
Dedekind cuts, one of the standard ways of constructing the reals from the rationals makes use of this. To construct $\sqrt 2$ you divide the positive rationals into two disjoint sets, those whose square is less than $2$ and those whose square is more than $2$.