VMO

a. It is obvious that $1\le \frac{2n^2+1}{n^2-n+1}\le 2$ for all $n\in {\mathbb{Z}}^{+}$. Hence, $0\le a_n\le \ln 2<1$ and $\lfloor a_n\rfloor =0$. It follows that $\{a_n\}=a_n$ and $$\underset{n\to \infty }{\lim }\{a_n\}=\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\ln \frac{2n^2+1}{n^2+n+1}=\ln 2.$$ Therefore, there exists a positive integer $n_0$ such that $$\{a_n\}>\ln 2-\frac{1}{2016},\forall n\ge n_0.$$ Then by direct calculation, it deduces that for $n>n_0$, $$\{a_n\}=a_n>\ln 2-\frac{1}{2016}>\frac{1}{2}.$$

b. $$\underset{n\to \infty }{\lim }(b_n-b_{n-1})=\underset{n\to \infty }{\lim }\ln \frac{(2n^2+1)(n^2+n+1)}{(2n^2-4n+3)(n^2-n+1)}=0.$$ Back to our problem, assume that there exists finite values of $n$ that $\{b_n\}<\frac{1}{2016}$. It follows that there exists a positive integer $n_0$ that $\{b_n\}\ge \frac{1}{2016}$ for all $n$. Because ${\lim }_{n\to \infty }(b_n-b_{n-1})=0$, it implies that there exists a positive integer $n_1$ that $$b_n-b_{n-1}<\frac{1}{2016}$$ for all $n\ge n_1$. Because $(b_n)$ is an increasing sequence and ${\lim }_{n\to \infty }{b}_n=+\infty$, it follows that there exists infinite values of $n>\max \{n_0,n_1\}$ which $[b_n]-[b_{n-1}]=1$. Consider such values of $n$, by the above inequality, we obtain $$[b_n]-[b_{n-1}]+\{b_n\}-\{b_{n-1}\}<\frac{1}{2016}$$ or $$\{b_{n-1}\}>\{b_n\}+\frac{2015}{2016}$$ Notice that $\{b_n\}\ge \frac{1}{2016}$, thus $\{b_{n-1}\}>1$. This contradiction finishes the given problem. $\square$