# Upper Bound for Lucas Number

## Theorem

Let $L_n$ denote the $n$th Lucas number.

Then:

$L_n < \paren {\dfrac 7 4}^n$

## Proof

The proof proceeds by complete induction.

For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:

$L_n < \paren {\dfrac 7 4}^n$

$\map P 1$ is the case:

 $\displaystyle L_1$ $=$ $\displaystyle 1$ $\displaystyle$ $<$ $\displaystyle \dfrac 7 4$

Thus $\map P 1$ is seen to hold.

### Basis for the Induction

$\map P 2$ is the case:

 $\displaystyle L_2$ $=$ $\displaystyle 3$ $\displaystyle$ $=$ $\displaystyle \dfrac {48} {16}$ $\displaystyle$ $<$ $\displaystyle \dfrac {49} {16}$ $\displaystyle$ $=$ $\displaystyle \paren {\dfrac 7 4}^2$

Thus $\map P 2$ is seen to hold.

This is the basis for the induction.

### Induction Hypothesis

Now it needs to be shown that if $\map P j$ is true, for all $j$ such that $0 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true.

This is the induction hypothesis:

$L_k < \paren {\dfrac 7 4}^k$

from which it is to be shown that:

$L_{k + 1} < \paren {\dfrac 7 4}^{k + 1}$

### Induction Step

This is the induction step:

 $\displaystyle L_{k + 1}$ $=$ $\displaystyle L_k + L_{k - 1}$ $\displaystyle$ $<$ $\displaystyle \paren {\dfrac 7 4}^k + \paren {\dfrac 7 4}^{k - 1}$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \paren {\dfrac 7 4}^{k - 1} \paren {1 + \dfrac 7 4}$ $\displaystyle$ $=$ $\displaystyle \paren {\dfrac 7 4}^{k - 1} \paren {\dfrac {11} 4}$ $\displaystyle$ $<$ $\displaystyle \paren {\dfrac 7 4}^{k - 1} \paren {\dfrac 7 4}^2$ $\displaystyle$ $=$ $\displaystyle \paren {\dfrac 7 4}^{k + 1}$

So $\map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction.

Therefore:

$\forall n \in \Z_{\ge 1}: L_n < \paren {\dfrac 7 4}^n$

$\blacksquare$