# Limit with fractional part of a rational number exists.

Given $$x\in\langle 1,\infty\rangle\cap\mathbb Q$$ Prove: $$\exists\lim_{n\to\infty}\{x^n\}\implies x\in\mathbb N$$ $$\{\cdot\}$$ is a fractional part of the number.

So far, even I couldn't prove it generally, I've used a calculator to notice that $$f(x_n)>f(x_{n-1})$$ and $$f(x_n)>f(x_{n+1})$$ if $$x$$ is not a natural number, where $$f(x_n)=\{x^n\}$$. That could mean $$f(x_n)$$ is not monotonous (maybe divergent) sequence, and maybe that leads to a contradiction. Yet, I couldn't come to any ideal solution though.

• Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments. Commented Feb 18, 2020 at 16:56
• If $x=k+r$ where $k$ is a positive integer and $0< r< 1$ then for any $\epsilon > 0$ there are infinite $n,m$ so that there is a integer $M < (k+r)^n < M+\epsilon$ and there is an integer $N$ so that $N-\epsilon < (k+r)^m < N$. Use archimedean principal. Commented Feb 19, 2020 at 1:30

Assume $$x=\frac{q}{p}, q>p>2, (q,p)=1$$ and let the limit of {$$x^n$$} be $$l$$. Pick $$\epsilon >0$$ very small such that $$100p(x+1)\epsilon < 1$$ say.

Then since {$$ml$$}, $$m\ge 1$$ integer is uniformly distributed modulo $$1$$ if $$l$$ irrational (edit later - here ud is overkill as noted in the comments, dense or accumulation at zero will do and that follows easily from Dirichlet-box arguments), or there is {$$ml$$}$$=0$$ if $$l$$ rational, and since obviously {$$mx^n$$} $$\to$${$$ml$$} for $$m$$ positive integer fixed, we can find $$m,N$$, s.t {$$mx^n$$} $$\le \epsilon, n \ge N$$.

But this means there are integers $$r_n \le mx^n, n \ge N$$, $$mx^n-r_n \le \epsilon, n \ge N$$. Multiplying the relation by $$x$$ and subtracting the relation for $$n+1$$ gives:

(edit per question in comments) - $$mx^{n+1}-r_nx \le x\epsilon$$ and $$mx^{n+1}-r_{n+1} \le \epsilon$$, so

$$|xr_n-r_{n+1}|=|(mx^{n+1}-r_nx)-(mx^{n+1}-r_{n+1})| \le |mx^{n+1}-r_nx|+|mx^{n+1}-r_{n+1}|$$, so

$$|xr_n-r_{n+1}| \le (x+1)\epsilon$$. However if $$xr_n-r_{n+1} \ne 0$$, we have $$|xr_n-r_{n+1}|=\frac{|qr_n-pr_{n+1}|}{p} \ge \frac{1}{p}$$, so $$p(x+1)\epsilon \ge 1$$ which contradicts the choice of $$\epsilon$$ above. Hence we must have $$xr_n=r_{n+1}, n \ge N$$.

This means $$qr_n=pr_{n+1}, n \ge N$$ so $$r_{N+1}=Aq$$ for some fixed integer $$A>0$$. But using $$qr_{N+1}=pr_{N+2}$$ we get that $$p$$ divides $$A$$ and $$r_{N+2}=\frac{q^2A}{p}$$

But now using $$qr_{N+2}=pr_{N+3}$$ we get $$p^2$$ divides $$A$$ and $$r_{N+3}=\frac{q^3A}{p^2}$$ so it is obvious that continuing this we eventually run out of powers of $$p$$ dividing $$A$$ so we get a contradiction! Done!

Note that the same proof applies when there are potentially finitely many limit points as we have joint uniform distribution (edit later - see comment above - dense will do) of the fractional parts of $$ml_1,..., ml_k$$ for rationally independent irrationals $$l_1,..l_k$$ and by appropriate integral multiplications we can reduce all the limit points to such and $$0$$ and find $$m$$ as before.

• I think you don't need uniformly distributed - dense will do. Commented Feb 19, 2020 at 5:07
• @marty - completely agree; ud is overkill, just accumulation at zero is needed and that is very easy with the usual Dirichlet-box type arguments Commented Feb 19, 2020 at 12:06
• @Conrad - it seems I didn't quite get how did you subtract the relation for n+1 Commented Feb 19, 2020 at 14:42
• edited to show it explicitly Commented Feb 19, 2020 at 14:50