Spring Mathematical Tournament

a) We shall prove the statement by induction if $a_1>\frac{19}{243}\left(\frac{1}{12}>\frac{19}{243}\right)$. It is true for $n\le 3$, since $a_2>\sqrt{3\cdot \frac{19}{243}+1}=\frac{10}{9}$ and $a_3>\sqrt{4\cdot \frac{10}{9}+1}=\frac{7}{3}$.

Assume that $a_n>n-\frac{2}{n}$ for some $n\ge 3$. Then $$a_{n+1}>\sqrt{(n+2)\left(n-\frac{2}{n}\right)+1}.$$ It is enough to show that right-hand side is bigger than $n+1-\frac{2}{n+1}$. It is easy to see that this is equivalent to $\frac{1}{2}>\frac{1}{n}+\frac{1}{(n+1{)}^2}$, which obviously holds for $n\ge 3$.

b) Note first that if $a_1=1$, then $a_n=n$ by induction, implying that $b_n=2^n\left(\frac{a_n}{n}-1\right)=0$. If $a_1<1$, then $a_n<n$ again by induction, i.e., $b_n<0$. We shall prove that $b_n<b_{n+1}$. This is equivalent to $\frac{a_n-n}{2n}<\frac{a_{n+1}-n-1}{n+1}$. The right-hand side equals $$\frac{a_{n+1}^2-n-1}{(n+1)(a_{n+1}+n+1)}=\frac{(n+2)a_n+1-(n+1{)}^2}{(n+1)(a_{n+1}+n+1)}=\frac{(n+2)(a_n-n)}{(n+1)(a_{n+1}+n+1)}.$$ It remains to show that $\frac{1}{2n}>\frac{n+2}{(n+1)(a_{n+1}+n+1)}$, i.e., $$(n+1)a_{n+1}>2n(n+2)-(n+1{)}^2=(n+1{)}^2-2,$$ which follows by a). The above arguments show that if $a_1>1$, then $b_n>b_{n+1}>0$. So the sequence $(b_n)$ is monotone and bounded; hence it converges.