China Mathematical Competition (Extra Test)

(1) According to the stated conditions we have $A_n(0,\frac{1}{n})$, and $B_n(b_n,\sqrt{2b_n})$ ($b_n>0$). From $|O{B}_n|=\frac{1}{n}$ we get $$b_n^2+2b_n={\left(\frac{1}{n}\right)}^2.$$ Thus, $$b_n=\sqrt{{\left(\frac{1}{n}\right)}^2+1}-1,n\in \mathbb{N}.$$

Since $2n^2{b}_n=1-n^2{b}_n^2>0$, we have $b_n+2=\frac{1}{n^2{b}_n}$ and $$\begin{array}{rl}{a}_n& =\frac{b_n(1+n\sqrt{2b_n})}{1-2n^2{b}_n}=\frac{b_n(1+n\sqrt{2b_n})}{1-(1-n^2{b}_n^2)}\\ & =\frac{1}{n^2{b}_n}+\frac{\sqrt{2}}{\sqrt{n^2{b}_n}}=b_n+2+\sqrt{2(b_n+2)}.\end{array}$$ Thus, $$a_n=\sqrt{{\left(\frac{1}{n}\right)}^2+1}+1+\sqrt{2\sqrt{{\left(\frac{1}{n}\right)}^2+1}+2}.$$ Since $\frac{1}{n}>\frac{1}{n+1}>0$, $a_n>a_{n+1}>4$ for any $n\in \mathbb{N}$.

(2) Let $c_n=1-\frac{b_{n+1}}{b_n}$ for $n\in \mathbb{N}$, then $$\begin{array}{rl}{c}_n& =\frac{\sqrt{{\left(\frac{1}{n}\right)}^2+1}-\sqrt{{\left(\frac{1}{n+1}\right)}^2+1}}{\sqrt{{\left(\frac{1}{n}\right)}^2+1}-1}\\ & =n^2\left[\frac{1}{n^2}-\frac{1}{(n+1{)}^2}\right]\cdot \frac{\sqrt{{\left(\frac{1}{n}\right)}^2+1}+1}{\sqrt{{\left(\frac{1}{n}\right)}^2+1}+\sqrt{{\left(\frac{1}{n+1}\right)}^2+1}}\\ & >\frac{2n+1}{(n+1{)}^2}\left[\frac{1}{2}+\frac{1}{2\sqrt{{\left(\frac{1}{n}\right)}^2+1}}\right]>\frac{2n+1}{2(n+1{)}^2}.\end{array}$$ Since $(2n+1)(n+2)-2(n+1{)}^2=n>0$, we obtain $$c_n>\frac{1}{n+2},n\in \mathbb{N}.$$ Let $S_n=c_1+c_2+\cdots +c_n,n\in \mathbb{N}$. If $n=2^k-2>1$ ($k\in \mathbb{N}$), then $$\begin{array}{rl}{S}_n& >\frac{1}{3}+\frac{1}{4}+\cdots +\frac{1}{2^k-1}+\frac{1}{2^k}\\ & =\left(\frac{1}{3}+\frac{1}{4}\right)+\left[\frac{1}{2^2+1}+\cdots +\frac{1}{2^3}\right]+\cdots +\left[\frac{1}{2^{k-1}+1}+\cdots +\frac{1}{2^k}\right]\\ & >2\cdot \frac{1}{2^2}+2^2\cdot \frac{1}{2^3}+\cdots +2^{k-1}\cdot \frac{1}{2^k}=\frac{k-1}{2}.\end{array}$$ Therefore, if we put $n_0=2^{4009}-2$, then for any $n>n_0$ we have $$\begin{array}{rl}\left[1-\frac{b_2}{b_1}\right]+\left[1-\frac{b_3}{b_2}\right]+\cdots +\left[1-\frac{b_{n+1}}{b_n}\right]& =S_n>S_{n_0}>\frac{4009-1}{2}\\ & =2004.\end{array}$$ Consequently, $$ \frac{b_2}{b_1} + \frac{b_3}{b_2} + \cdots + \frac{b_n}{b_{n-1}} + \frac{b_{n+1}}{b_n} < n - 2004, \quad n > n_0.