Prove that $\forall n, \, \exists N,x :\lfloor{x^{N}}\rfloor =n \, \land \,\lfloor{x^{N+1}}\rfloor =n+1$

Question

The question is related to the interesting problem raised by following OP.

Notably I'm trying to prove the following fact

For any $n\in \mathbb{N} \quad \exists x\in \mathbb{R} \quad x>1$ and $\exists N\in \mathbb{N}$ such that

$$\lfloor{x^{N}}\rfloor =n \quad \land \quad \lfloor{x^{N+1}}\rfloor =n+1$$

My idea for the proof is to consider

• $x=\sqrt[N]{n} \implies \lfloor{x^{N}}\rfloor = \lfloor{{(\sqrt[N]{n})}^{N}}\rfloor =\lfloor n\rfloor=n$

and then we need that

• $x^{N+1}=x\cdot n=n+1+\alpha\,$ for some $\alpha \in(0,1)$

therefore finally we need to show that for any $n\in \mathbb{N} \quad \exists \alpha\in \mathbb{R} \quad \alpha \in(0,1)$ and $\exists N\in \mathbb{N}$ such that

$$\sqrt[N]{n}=1+\frac{1+\alpha}{n}$$

but I'm totally stuck here and I can't find any method to prove that.

Thanks in advance for any idea or suggestion about that!

Answer 1

Start two sequences $(x_i)$ and $(N_i)$ with $x_0=n, N_0=1$ and define $x_{i+1}=\sqrt{x_i}, N_{i+1}=2N_i, i=0,1,\ldots$. Obviously we have $x_i^{N_i}=n$ and $\lim_{i \to \infty} x_i=1$. So for some $k$, $x_k < \frac{n+2}{n+1}$.

We consider $x_k^{N_k+0}, \ldots, x_k^{N_k+i}, \ldots$. Since $x_k > 1$, there is an index $r \ge 0$ with $x_k^{N_k+r} < n+1$ and $x_k^{N_k+r+1} \ge n+1$. Since $x_k < \frac{n+2}{n+1}$, we also have $x_k^{N_k+r+1} = x_kx_k^{N_k+r} < (n+1)\frac{n+2}{n+1} = n+2$ which finally leads to the desired

$$ \lfloor x_k^{N_k+r}\rfloor=n, \quad \lfloor x_k^{N_k+r+1}\rfloor = n+1$$

Answer 2

This is long after you received the excellent answer by Ingix(+1), but I'd like to present an arguably more elementary solution.

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We'll use $x = n^{1/N}$ as in your Q. We need to argue that $\exists N \in \mathbb{N}:$ \begin{align} n+1 &\le n^{1 + 1/N} < n+2 \\ \iff \frac{\ln (n+1)}{\ln n} &\le 1 + \frac{1}{N} < \frac{\ln (n+2)}{\ln n} \\ \iff \frac{\ln(1+1/n)}{\ln n} &\le \frac{1}{N} < \frac{\ln (1 + 2/n)}{\ln n} \\ \iff \frac{\ln n }{\ln(1 + 2/n)} &< N \le \frac{\ln n}{\ln( 1 + 1/n)} \end{align}

But this is true so long as the difference between the left and right hand side above is at least $1$. We can use the the following elementary inequality: $$\mathrm{for}\, x\in (0,1): \quad 2x/3 \le \ln(1+x) \le x \iff \frac{1}{x} \le \frac{1}{\ln(1+x)} \le \frac{3}{2x}. $$

Thus, $ \frac{\ln n}{\ln(1 + 1/n)} \ge n \ln n$ and $ \frac{\ln n}{\ln(1 + 2/n)} \le \frac{3n}{4} \ln n$. The gap between them is at least $ \frac{n}{4} \ln n $ which is seen to be $> 1$ for $n\ge 4.$

For $n = 1,2,3$ witnesses are found easily: $x = 4/3$ and $N = 2,3,4$ respectively.