The inequality about recurrence sequence Sequence $(x_n)$ is difined
$x_1=\frac {1}{100}, x_n=-{x_{n-1}}^2+2x_{n-1}, n\ge2$
Prove that
$$\sum_{n=1}^\infty [(x_{n+1}-x_n)^2+(x_{n+1}-x_n)(x_{n+2}-x_{n+1})]\lt \frac {1}{3} $$
I found relation $(1-x_n)=(1-x_{n-1})^2$
I don't know what to do next.
There is a real number which is less than $\frac {1}{3}$?
I need your help. 

A direct proof (note that I've shifted indices from starting at 1 to 0):
First, notice that $x_n\to1$ is the only possible limit. ($x=-x^2+2x \implies (x-1)^2=0$)
[edit]
The obvious mistake in my algebra was pointed out -- $x^2=x$ so $x=0$ or $1$. Recentering about $x=0$ doesn't change the recurrence, but recentering about $x=1$, as you can see below, yields a relation that can be solved by inspection.[/edit]
When in doubt, recenter your system about the fixed point; define $a_n=1-x_n$. Then,
$$ a_{n+1} = 1-x_{n+1} = 1-[-x_n^2+2x_n] = 1+x_n^2-2x_n=(1-x_n)^2=a_n^2. $$
This allows us to solve the system exactly; $a_n = a_0^{2^n}$, or $1-x_n=(1-x_0)^{2^n}$.
It should now be a matter of algebra to evaluate the sum.
