sorry, not sure which subforum this should go to.

Sequence u1, u2, u3,... is know to converge to the limit m and is such that $\displaystyle u_{n+1}=\frac{9u_n+3}{u_n+7}$

a) find the value of m.

b) it is given that $\displaystyle u_n=1$. by first writing $\displaystyle u_n$ in terms of $\displaystyle u_{n+1}$, and considering $\displaystyle 3-u_n$, prove by induction that $\displaystyle u_n<3$ for all positive integers n.

for a.

i substitute m, giving $\displaystyle m=\frac{9m+3}{m+7} $

solving this gives $\displaystyle m=-1,3$

how do i immediately know whether it is -1 or 3 that it converges to? i know i can draw a graph of $\displaystyle u_{n+1}$ against $\displaystyle u_n$, and draw a line of y=x, and based on the gradient of the curve determine which it converges to. but i believe im not supposed to do that, this topic is only induction. and also this is a short question which probably carries very little marks. there must be some direct and quick way to immediately see which it converges to?

for b.

i proved that it is valid for n=1.

i assumed that it is valid for n=k.

that gives $\displaystyle u_k<3$

expressing $\displaystyle u_n$ in terms of $\displaystyle u_{n+1}$, $\displaystyle u_n=-\frac{7u_{n+1}-3}{u_{n+1}-9}$

considering $\displaystyle 3-u_n$,

$\displaystyle 3-u_n=\frac{10u_{n+1}-30}{u_{n+1}-9}$

since it is assumed $\displaystyle u_k<3$,

then $\displaystyle \frac{10u_{n+1}-30}{u_{n+1}-9}>0$

this is where my problem starts. i can't show from here that $\displaystyle u_{n+1}<3$

how? in fact what i get is $\displaystyle u_{n+1}<3,u_{n+1}>9$. theres this $\displaystyle u_{n+1}>9$, which shouldn't be if i were to prove this.