This question comes from Rational Points on Elliptic Curves (Silverman & Tate) exercise 3.3(a).
$3.3$. Let $C$ be a rational cubic curve given by the usual Weierstrass equation. Prove that for any rational point $P \in C(\mathbb{Q})$, the limit $$\hat{h}(P)=\lim _{n \rightarrow \infty} \frac{1}{4^n} h\left(2^n P\right)$$ exists. The quantity $\hat{h}(P)$ is called the canonical height of $P$.
Note that the height of $x$ is $H(x)=H\left(\frac{m}{n}\right)=\max \{|m|,|n|\}$ and $h(P)=\log H(P)$
It gave a hint to suggest showing the sequence is Cauchy, so here is my attempt:
We wish to show that, given $\varepsilon>0$ there exists $N$ such that if $m, n > N$ then $$\left|4^{-n}h(2^nP) - 4^{-m}h(2^mP)\right|<\varepsilon.$$ Write $P = x/y$ so we have
$$\left|4^{-n}h(2^n\frac x y) - 4^{-m}h(2^m\frac x y)\right|.$$
Now, I'm not sure how to proceed but my best idea is that because $e^{\varepsilon}$ can be made arbitrarily small just like $\varepsilon$ itself, we can raise everything to the power of $e$ and get rid of small $h$. Is this the right direction?
I will go ahead and answer my question since I now see how to do it.
We will show that the sequence is Cauchy. We (I) have already shown that there exists a constant $C$ such that for all $Q \in E(\bar{K})$, $$\left|h(2 Q)-4 h(Q)\right| \leq C.$$
Now we will show that the series is a telescoping sum (how clever!). For integers $N \geqslant M \geqslant 0$ we have
$$ \begin{align*} \mid 4^{-N} h\left(2^N P\right) - 4^{-M} h({\left.2^M P\right) \mid} &= \left|\sum_{n = M}^{N-1} 4^{-n - 1} h\left(2^{n+1} P\right) - 4^{-n} h\left(2^n P\right)\right| \\ &\leqslant \sum_{n=M}^{N-1} 4^{-n-1}\left|h\left(2^{n+1} P\right) - 4 h\left(2^n P\right)\right| \\ & \leqslant \sum_{n=M}^{N-1} 4^{-n-1} C \quad \text { taking } Q=2^n P \\ & \leqslant 4^{-M} C. \end{align*}$$ Therefore, the sequence $4^{-N} h\left(2^N P\right)$ is Cauchy, and thus the limit exists!
Thanks to user KCd for pointing me in the right direction.