<s>[INST] Evaluating $\liminf_{n\to\infty}n\{n\sqrt2\}$ 
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 $0.36$.
 [/INST] 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 $\displaystyle\;\liminf_{n\to\infty}\, n\{n\sqrt{2}\} = \frac{1}{2\sqrt{2}}$
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