The series is given as $a_{n+1}=\frac{8a_{n}-3}{2a_{n}+1}$ and $a_{1}=1$
I have no idea how to prove if this series is monotone or not.
I have tried rearranging the equation to form $a_{n+1}>a_{n}$ but nothing seems to work.
Please if you know how to solve this it would help a ton!
Just for your curiosity.
In my answer to this question, I detailed the steps for the solution of first-order rational difference equation (have a look here) $$a_{n+1} = \frac{m\,a_n + x}{a_n + y} $$
With your parameters $m=4$, $x=-\frac 32$ and $y=\frac 12$, the results for $a_n$ are extremely simple.
On the other side, do not forget that, if the limit $L$ exists, it is the solution of $L=\frac{8 L-3}{2 L+1}$.
$$a_{n+1}-3=\frac{2(a_n-3)}{2a_n+1}\tag1$$ $$2a_{n+1} - 1 = \frac{7(2a_n-1)}{2a_n+1} \tag 2$$
Since $a_1=1 > \frac 12>0$, from $(2)$ we know $2a_n-1>0, \forall n$. We also know $a_n < 3, \forall n$ from $(1)$.
Therefore $$a_{n+1}-a_n = \frac{(3-a_n)(2a_n-1)}{2a_n+1}>0.\blacksquare$$
Remark 1: We derive $(1)$ and $(2)$ based on the solutions of the characteristic equation $\lambda = \frac{8\lambda - 3}{2\lambda + 1}$ which are $3$ and $\frac 12$.
Remark 2: We can derive the closed form solution for $a_n$ easily as follows
Divide $(1)$ by $(2)$ and we obtain
$$\frac{a_{n+1}-3}{2a_{n+1}-1} = \frac 27 \cdot \frac{a_n-3}{2a_n-1} \tag 3$$
Now $(3)$ is a geometric sequence, $$ \frac{a_n-3}{2a_n-1} = \left( \frac 27 \right)^{n-1} \cdot \frac{8a_1-3}{2a_1+1} $$ and the rest is straightforward.
