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 $0.36$.
 A: 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 $\displaystyle\;\liminf_{n\to\infty}\, n\{n\sqrt{2}\} = \frac{1}{2\sqrt{2}}$
A: 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,
$$
\liminf_{n\to\infty}n\left\{n\sqrt2\right\}\le\frac1{2\sqrt2}
$$

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

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