$$\frac{1}{2n}-\frac{1}{2n+3}$$
Can I separate each part before and after subtraction and determine their montonicity? How does that relate to the combination?
$$a_n = \frac{1}{2n}$$
$$b_n = \frac{1}{2n+3}$$
both individually should be bounded and monotonic but is there a theorem such that subtracting two bounded and monotonic sequences means the difference will also be bounded and be monotonic?
On
A sequence of real numbers that is bounded above (below) and is monotone increasing (decreasing) is convergent. The sum and difference of convergent sequences is of course convergent. However, I think it is easier to see here that $$ \frac{1}{2n} - \frac1{2n+3} = \frac{3}{2n(2n+3)}\sim \frac{1}{n^2}\stackrel{n\to\infty}\longrightarrow0. $$
To show that it's monotone decreasing, simply use induction and the fact that $a_n-b_n = \frac{3}{2n(2n+3)}$ and induction. It's trivial to show that it is bounded since $a_n-b_n >0\;\forall n.$ By the Monotone Convergence Theorem, it has a limit.