Consider, $f(x_n )= x_{n+1}$. so see $f(x_n)-f(x_{n-1}) >0$. So it is increasing .now think that $ \lim_{n\to \infty} \frac{x_{n+1}}{x_n} =1 $. So it can diverge .Now see that. $\lim_{n\to \infty}\frac{x_n}{n}=\lim_{n\to \infty} \frac{x_{n+1}-x_n}{1}=\lim _{n\to \infty}x_n^{1/3}$ stolze Cesaro says these.....,... ....... Now look $\frac {x_n}{n^b} = \frac {x_n^{b/3}}{x_n ^{b-1}}$ ...by stolze Cesaro again .... So,we can say $\frac{b}{3}=b-1$. I.e $ b=\frac{3}{2} $. and hence $a= 1$. So,$\boxed{b=\frac{3}{2},a=1}$ QED $\square$.