If $u_1=4$ and $u_{n+1}=\frac{3u_n+5}{u_n-1}$, then $u_n$ approaches $a=5$ as $n\to\infty$. Show that, if $u_n>a$, then $u_{n+1}<u_n$.

Question

The sequence of positives numbers $u_1,u_2,u_3...$ is such that $u_1=4$ and $u_{n+1}=\dfrac{3u_n+5}{u_n-1}$ for all $n\ge1$.

1. Given that $u_n\rightarrow a$ as $n\rightarrow\infty$, find the value of $a$. (solved) Answer: $a=5$.

2. If $u_n\gt a$, show that $u_{n+1}\lt u_n$. (This is the part where I'm stuck.)

Answer 1

$$u_n-u_{n+1} = u_n-\frac{3u_n+5}{u_n-1} = \frac{u_n^2-4u_n-5}{u_n-1}$$

$$ = \frac{(u_n-5)(u_n+1)}{u_n-1}>0$$

Since $u_n-5$, $u_n+1,$ and $u_n-1$ are positive for $u_n>a=5$ we are done.