natural sequence I'm trying to solve this exercise on sequences :
$$u_{n+1}=\frac{3u_n+4}{2u_n+3}$$ and $$u_0=1$$ for any natural number 
1)Find a and b as $$u_{n+1}=a+\frac{b}{2u_n+3}$$
I've found a=3/2 and b=-1/2.
2) Show that $$0\leq u_n \leq 3/2$$ and $$u_{n+1}-\sqrt{2}=\frac{(u_n-\sqrt{2})(3-2\sqrt{2})}{2u_n+3}$$
Was not a problem.
3)Show that the real number $$c=\frac{3-2\sqrt{2}}{3} $$ satisfies $$0<c<1$$
4) However I do not succeed to solve the last two questions : I have to show that
$${\mid u_{n+1}-\sqrt{2}\mid}\leq c\mid u_n-{\sqrt{2}}\mid $$ and $${\mid u_{n}-\sqrt{2}\mid}\leq c^n\mid u_0-{\sqrt{2}}\mid $$ Do you have any ideas how to proove it ?
Thank you for your help ;-)

You know that:
$${\mid u_{n+1}-\sqrt{2}\mid}\leq {\mid \frac{(u_n-\sqrt{2})(3-2\sqrt{2})}{2u_n+3}\mid}$$
Now 
$${\mid \frac{(u_n-\sqrt{2})(3-2\sqrt{2})}{2u_n+3}\mid}\leq c\mid u_n-{\sqrt{2}}\mid \Leftrightarrow\\
{\mid \frac{(3-2\sqrt{2})}{2u_n+3}\mid}\leq c  \Leftrightarrow\\
 3 \leq {\mid 2u_n+3\mid}$$
Why is the last inequality true? 
The last claim follows now by induction on $n$.
