Evaluating $\liminf_{n\to\infty}n\{n\sqrt2\}$

Question

How can we evaluate $$\liminf_{n\to\infty}n\{n\sqrt2\},$$where $\{\cdot\}$ denotes the fractional part of $\cdot$?

The first thing came to my mind is Pell's equation $x^2-2y^2=1$.
Knowing that $\sqrt2$ has a continued fraction $[1;2,2,2\cdots]$, I tried to estimate the limit by using $\sqrt2$'s convergents' denominators. It seems like the limit approximately equals to $0.36$.

Answer 1

For $n > 0$, let $m = \lfloor n\sqrt{2}\rfloor$. Since $\sqrt{2} \not\in \mathbb{Q}$,

$$n\sqrt{2} > m \implies 2n^2 > m^2 \implies 2n^2 \ge m^2 + 1 \implies \sqrt{2}n \ge \sqrt{m^2+1}$$

This implies $$n\{n\sqrt{2}\} \ge n(\sqrt{m^2+1} - m) \ge \frac{1}{\sqrt{2}}\sqrt{m^2+1}(\sqrt{m^2+1}-m)\\ = \frac{1}{\sqrt{2}}\frac{\sqrt{m^2+1}}{\sqrt{m^2+1}+m} \ge \frac{1}{2\sqrt{2}}$$ As a result, $\displaystyle\;\liminf_{n\to\infty}\, n\{n\sqrt{2}\} \ge \frac{1}{2\sqrt{2}}$.

For the other direction, consider following pair of sequences of integers $(m_k), (n_k)$ defined by

$$m_k + n_k \sqrt{2} = (1 + \sqrt{2})^{2k+1}\quad\text{ for } k \in \mathbb{N}$$

It is easy to check they are increasing and $m_k^2 - 2n_k^2 = -1$. As a result,

$$\liminf_{n\to\infty}\, n\{n\sqrt{2}\} \le \liminf_{k\to\infty}\, n_k\{n_k\sqrt{2}\} = \liminf_{k\to\infty}\frac{1}{\sqrt{2}}\frac{\sqrt{m_k^2+1}}{\sqrt{m_k^2+1} + m_k}\\ = \lim_{m\to\infty} \frac{1}{\sqrt{2}}\frac{\sqrt{m^2+1}}{\sqrt{m^2+1}+m} = \frac{1}{2\sqrt{2}}$$

Combine these, we get $\displaystyle\;\liminf_{n\to\infty}\, n\{n\sqrt{2}\} = \frac{1}{2\sqrt{2}}$

Answer 2

Using continued fractions, we get $$ \begin{align} p_n &=\frac{\left(\sqrt2+1\right)^{2n-1}-\left(\sqrt2-1\right)^{2n-1}}{2}\\ &=\sum_{k=0}^{2n-1}\binom{2n-1}{k}\sqrt2^{2n-1-k}\frac{1^k-(-1)^k}2\\ &=\sum_{k=1}^n\binom{2n-1}{2k-1}2^{n-k}\\[12pt] &\in\mathbb{Z} \end{align} $$ and $$ \begin{align} q_n &=\frac{\left(\sqrt2+1\right)^{2n-1}+\left(\sqrt2-1\right)^{2n-1}}{2\sqrt2}\\ &=\sum_{k=0}^{2n-1}\binom{2n-1}{k}\sqrt2^{2n-1-k}\frac{1^k+(-1)^k}{2\sqrt2}\\ &=\sum_{k=0}^{n-1}\binom{2n-1}{2k}2^{n-k-1}\\[12pt] &\in\mathbb{Z} \end{align} $$ which give $$ q_n\sqrt2-p_n=\left(\sqrt2-1\right)^{2n-1} $$ and $$ q_n\overbrace{\left(q_n\sqrt2-p_n\right)}^{\left\{q_n\sqrt2\right\}}=\frac{1+\left(3-2\sqrt2\right)^{2n-1}}{2\sqrt2} $$ Thus, $$ \liminf_{n\to\infty}n\left\{n\sqrt2\right\}\le\frac1{2\sqrt2} $$

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For any $p,q\in\mathbb{Z}$ so that $p\lt q\sqrt2\lt p+1$, we have $$ \begin{align} 1 &\le2q^2-p^2\\ &=\left(q\sqrt2-p\right)\left(q\sqrt2+p\right) \end{align} $$ giving $$ \begin{align} q\left(q\sqrt2-p\right) &\ge\frac{q}{q\sqrt2+p}\\ &\ge\frac{q}{2q\sqrt2-1} \end{align} $$ Therefore, $$ \liminf_{n\to\infty}n\left\{n\sqrt2\right\}\ge\frac1{2\sqrt2} $$

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Thus, $$ \liminf_{n\to\infty}n\left\{n\sqrt2\right\}=\frac1{2\sqrt2} $$