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 $$ \lim_{n \to \infty} \vert L(n)  \phi^n \vert = 0? $$
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Yes, you can solve the recurrence for $L_n$ and you will get something in terms of $\phi.$ See wikipedia. – Phicar Nov 14 '16 at 01:01

1There is a very easy reason for this. The closed form for these numbers is something like $\phi^n\phi^{n}$ (probably says this in the Wikipedia article). – Matt Samuel Nov 14 '16 at 01:11
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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) $
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