How do you solve this recurrence ? I only know how to solve it if the recurrence is liniar .Can you give me a link or can you post a solution ? Thank you

$x_{n+1}=x_{n}(x_{n}+1)$, therefore $\frac{1}{x_{n}+1}=\frac{1}{x_{n}}-\frac{1}{x_{n+1}}$. It give $\sum_{k=1}^{n}\frac{1}{x_{k}+1}=\frac{1}{x_{1}}-\frac{1}{x_{n+1}}$. Because $x_{1}=\frac{1}{3}, x_{2}=x_{1}(1+x_{1})=\frac{4}{9}>x_{1}, x_{3}=\frac{52}{81},x_{4}=\frac{6916}{6561}>1 \ x_{2008}>2^{2004}$. It give $[3-\frac{1}{x_{2008}}]=2.$

how do you show x_2008>2^2004

drapt@ wrote: how do you show x_2008>2^2004 sufficiently $x_{2008}>1$. From $x_{4}>1$ we have $x_{5}>2,x_{6}>x_{5}^{2}>2^{2^{1}}$, therefore $x_{2008}>2^{2^{2003}}.$

but how can you say that if it holds for x^4 , x^5 ,x^6 it too wud hold for x^2008

i got the actual explanation , since x_n+1 - x_n > 0 and x_4 >1 therefore definately, x_2008>1 i i.e 0<1/x_2008<1 or 2<3- 1/x_2008<3, therfore its integral part is 2.