Is there a proof for why the difference between the n-th power of phi and the n-th Lucas number converges to zero?

Question

Let $\epsilon(n)$ be the absolute value of the difference between the $n$th Lucas number ($L(n)$) and the $n$th power of $\phi$. $\epsilon(n)$ pretty clearly converges to zero, and does so pretty fast. But is there a proof for $$ \lim_{n \to \infty} \vert L(n) - \phi^n \vert = 0? $$

Answer 1

The closed form for Lucas' number can be easily obtained by writing the recurrence relation and solving it by assuming a solution of form $a\cdot\alpha^n + b\cdot\beta^{-n}$ and substituting $n=1,2,3,4$ to get $4$ equations to solve 4 unknowns. It is given by:-

$$L_n = \phi^n+(1-\phi)^{n}$$

Therefore, $$\lim_{n\to\infty} |L_n-\phi^n| = \lim_{n\to\infty} |(1-\phi)^{n}| = 0$$

(because $|1-\phi|=\frac{\sqrt{5}-1}{2}<1) $