# 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 $u_{n+1}<u_n$.

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.)

$$u_n-u_{n+1} = u_n-\frac{3u_n+5}{u_n-1} = \frac{u_n^2-4u_n-5}{u_n-1}$$
$$= \frac{(u_n-5)(u_n+1)}{u_n-1}>0$$
Since $$u_n-5$$, $$u_n+1,$$ and $$u_n-1$$ are positive for $$u_n>a=5$$ we are done.