Find the limit of $\frac{T_n}{5n+4}$ Given that $U_0=0$, $U_{n+1}=\frac{U_n+3}{5-U_n}$

*

*Find the limit of $U_n$

*Set $T_n= \sum_{k=1}^n \frac1{U_k-3}$, find $\lim_{n\to+\infty}\frac{T_n}{5n+4}$
Approaches
So for the first question, I think it is flat out easy, because I can take:
\begin{align}
U_{n+1}-1=\frac{2U_n-2}{5-U_n} \\
U_{n+1}-3=\frac{4U_n-12}{5-U_n}
\end{align}
And thus
\begin{align}
\frac{U_{n+1}-1}{U_{n+1}-3}=\frac12 \times \frac{U_n-1}{U_n-3}
\end{align}
And thus converges to $0$
Now for the second question, I can't really tell what to do. My initial thought was to use telescoping:
$5U_{n+1}-U_n-3=U{n+1}-U_n$, but apparently the number 5 in front of $U_{n+1}$ is hindering the process.
Any help is appreciated!
 A: Part 1
Your method works just fine, I just wanted to use a different method to illustrate a technique similar to what I will use in Part 2.
Assuming we know that $U_n$ converges (which is the premise of the question), we see that $$\lim_{n\to\infty}U_{n+1}=\lim_{n\to\infty}U_{n}.$$ Then, $$\lim_{n\to\infty}\frac{U_n+3}{5-U_n}=\lim_{n\to\infty}U_n.$$ Then, in the limit, $$\frac{U_n+3}{5-U_n}=U_n,$$ and solving for $U_n$ gives $U_n=1$ or $U_n=3.$ However, since $U_0=0,$ it is clear that $U_n$ cannot exceed $1,$ and hence, $\lim_{n\to\infty}U_n=1$ (aside: if $U_0<3,$ then the limit tends to $1$, and tends to $3$ otherwise).
Part 2
Again, the premise of the question assumes that $\lim_{n\to\infty}\frac{T_n}{5n+4}$ converges. Then, we can write $$\lim_{n\to\infty}\frac{T_{n+1}}{5(n+1)+4}=\lim_{n\to\infty}\frac{T_n}{5n+4}.$$ Further, we see that, through the sum, $$T_{n+1}=\sum_{k=1}^{n+1} \frac{1}{U_k-3}=\left(\sum_{k=1}^{n} \frac{1}{U_k-3}\right)+\frac{1}{U_{n+1}-3}=T_n+\frac{1}{U_{n+1}-3}.$$ We also know that $\lim_{n\to\infty} U_{n+1}=1,$ so $\lim_{n\to\infty} \frac{1}{U_{n+1}-3}=-\frac{1}{2}.$ Thus, in the limit, we have $$\frac{T_{n+1}}{5(n+1)+4}=\frac{T_n-\frac{1}{2}}{5n+9}=\frac{T_n}{5n+4}.$$ Finally, solving for $T_n$ gives $$T_n=-\frac{1}{10}(5n+4),$$ so $$\lim_{n\to\infty}\frac{T_n}{5n+4}=-\frac{1}{10}.$$
