BMO 2010 Shortlist

For every $n\in N^{\ast }$ from the given recurrence relation we have $$a_{n+1}-a_n=(a_n-1{)}^2.$$ So, the sequence $(a_n)$ is increasing and therefore, since $a_1=2>1$, we have for every $n\in N^{\ast }$ $$a_{n+1}>a_n>1.$$ From the given recurrence relation for every $n\ge 2$ we obtain the equalities $$a_n-1=a_{n-1}(a_{n-1}-1),$$ $$a_{n-1}-1=a_{n-2}(a_{n-2}-1),$$ $⋮$ $$a_2-1=a_1(a_1-1).$$ $$a_n=1+a_1{a}_2\cdots a_{n-1}.$$ Dividing both sides of the last relation by $a_1{a}_2\cdots a_{n-1}{a}_n$ we get $$\frac{1}{a_1{a}_2\cdots a_{n-1}}=\frac{1}{a_1{a}_2\cdots a_{n-1}{a}_n}+\frac{1}{a_n}$$ or $$\frac{1}{a_n}=\frac{1}{a_1{a}_2\cdots a_{n-1}}-\frac{1}{a_1{a}_2\cdots a_{n-1}{a}_n}(1)$$ We put $n=2,3,4,\dots ,k$ in (1) and adding recursively the relations we get $$\sum _{i=2}^k\frac{1}{a_i}=\frac{1}{a_1}-\frac{1}{a_1{a}_2\cdots a_{k-1}{a}_k}\iff \sum _{i=1}^k\frac{1}{a_i}=1-\frac{1}{a_1{a}_2\cdots a_{k-1}{a}_k}<1(2)$$ for every $k\in N^{\ast }$, since $a_i>1$ for every $i\in N^{\ast }$. For every $n\in N^{\ast }$, $n\ge 2$, from the hypothesis we have $a_n-a_{n-1}=(a_{n-1}-1{)}^2\ge 1$. Therefore, inductively we get $$a_n\ge n-1+2=n+1>n.$$ So, for every $k\in N^{\ast }$ we have $$0<\frac{1}{a_1{a}_2\cdots a_{k-1}{a}_k}<\frac{1}{a_k}<\frac{1}{k}$$ and $$1-\frac{1}{a_1{a}_2\cdots a_{k-1}{a}_k}>1-\frac{1}{k}(3)$$ For every $0\le \alpha <1$ we'll find $k\in N^{\ast }$ such that $$\sum _{i=1}^k\frac{1}{a_i}>\alpha .(4)$$ For if $$1-\frac{1}{a_i}>\alpha \iff k-1>k\alpha \iff k>\frac{1}{1-\alpha }\iff k\ge \lfloor \frac{1}{1-\alpha }\rfloor +1,(5)$$ where $[x]$ denote the largest integer less than or equal to $x$. From the relations (2),(3),(5) we obtain (4). So, $L=1$. $\square$