Let $\{x_n\}$ be a sequence of real numbers and $\left| x_{n+1}-x_n \right| \lt \left(\frac{3}{4}\right)^n \ \forall n \in \mathbb{N}$.
My strategy is to show the sequence is Cauchy and since it's a sequence in $\mathbb{R}$, it is convergent.
Proof:
Let $m,n \in \mathbb{N}$ and w.l.o.g. let $n \gt m$, then
$\left| x_n - x_m \right|$
$= \left| x_n - x_{n-1} + x_{n-1} - x_{n-2} + x_{n-2} + ...+ x_{m+1} - x_m \right|$
$\leq \left| x_n - x_{n-1} \right| + ... + \left| x_{m+1} - x_m \right|$
$\lt \left(\frac{3}{4}\right)^{n-1} + \left(\frac{3}{4}\right)^{n-2} + ... + \left(\frac{3}{4}\right)^m = \left(\frac{3}{4}\right)^m \left[ \left(\frac{3}{4}\right)^{n-m-1} + ... + 1 \right]$ ......$(\dagger)$
Using the geometric formula $\left[ \left(\frac{3}{4}\right)^{n-m-1} + ... + 1 \right] = 4 \left( 1 - \left( \frac{3}{4} \right)^{n-m-1} \right) \lt 4$
So $\dagger$ becomes,
$\left| x_n - x_m \right| \lt 4 \left( \frac{3}{4} \right)^m$
Now we want $4 \left( \frac{3}{4} \right)^m \lt \epsilon$
or $\left( \frac{3}{4} \right)^m \lt \frac{\epsilon}{4}$
Can I simply say, choose $N \in \mathbb{N}$ such that for $m \geq N$ the above is true?
Can I get some verification of my approach as well as some help with what to do in the last step?
Concerning the last step, all you need is to take $N\in\mathbb N$ such that $\left(\frac34\right)^N<\frac\varepsilon4$. Then, if $n>m\geqslant N$,$$\lvert x_m-x_n\rvert<\left(\frac34\right)^m\leqslant\left(\frac34\right)^N<\frac\varepsilon4.$$The rest of what you did looks fine.