Lucas sequence inequality

Question

It is a common exercise to prove that $$a_n \lt {\left(\frac{7}{4}\right)}^{n}$$
where

$a_n$ is the $n^{th}$ term of the sequence
$a_n = a_{n-1} + a_{n-2}$
$a_1 = 1$
$a_2 = 3$

It can be proved by using mathematical induction pretty quickly. But my question is regarding the right-hand side of the inequality

Q1. Does the number $(\frac{7}{4})^n$ have any special significance?

By special significance, I mean that is there any reason why this specific number has been chosen for this inequality? I couldn't find anything special about this number in this inequality. I tried plugging in different numbers but my effort tells me that $\frac{7}{4}$ may be the smallest number satisfying this inequality. If this is so, it brings me to the next question

Q2. How are we arriving at this number?

Was this number the result of "hit-and-trial" or is it possible to arrive at this exact number by other methods?

Any insight will be appreciated. Thanks in advance for your answer!

Answer 1

$(a_n)$ is the sequence of Lucas numbers, therefore I'll use the common notation $L_n$ instead. The known closed formula $$ L_n = \left( \frac{1+\sqrt 5}{2}\right)^n + \left( \frac{1-\sqrt 5}{2}\right)^n $$ shows that $(L_n)$ grows exponentially: $$ L_n \sim \left( \frac{1+\sqrt 5}{2}\right)^n \text{ for } n \to \infty \, . $$ It follows that there is some $q > 0$ such that $$ \tag{$*$} L_n < q^n \text{ for all } n =1, 2, 3, \ldots \, . $$

If $q > 0$ satisfies $(*)$ then necessarily $3 = L_2 < q^2$, i.e. $q > \sqrt 3$. On the other hand it is not difficult to show with induction that if $q > \sqrt 3$ then $(*)$ is satisfied. For the inductive step one needs that $1 + q < q^2$.

So $(*)$ is satisfied if and only if $q > \sqrt 3 \approx 1.732$. Your choice $q=7/4$ is a rational number satisfying that condition, but it is not best possible.

The question remains: How does one come of with $7/4$ as a “good” rational approximation of $\sqrt 3$?

A systematic approach is that of continued fractions. The “successive convergents” of a continued fraction of a real number $x$ give a sequence of rational numbers which converge quickly to $x$ and are alternatingly smaller and larger than $x$.

For $x=\sqrt 3$ the convergents which are larger than $\sqrt 3$ are $$ 2, 7/4 = 1.75, 26/15 \approx 1.73333, 97/56 \approx 1.73214, \ldots. $$

So your choice $q=7/4$ is a rational number larger than $\sqrt 3$, and it can be interpreted as a continued fraction approximation.

Answer 2

Short answer: they picked $\frac{7}{4}$ for a reason.

However, my take is that there is nothing interesting encoded in the fraction $\frac{7}{4}$. The function $f(n) = (\frac{7}{4})^n$ serves as a modest cap for the $g(n) = a_n$ where n is integer.

What IS interesting is the inherent nature of Lucas sequence. You can explore generating functions, relationship to Fibonnaci sequence, etc. You may find this webpage interesting. Two of the things pointed out in this web page is that:

For every series formed by adding the latest two values to get the next, and no matter what two positive values we start with the ratio between two successive terms will eventually approach phi=1·6180339..

For example:

$\frac{3}{1}=3 \quad \frac{4}{3}=1.333 . . \quad \frac{7}{4}=1.75 \quad \frac{11}{7}=1.5714 . . \quad \frac{18}{11}=1.6363 . . \quad \frac{29}{18}=1.6111 .$.

You asked if there's anything special about $\frac{7}{4}$. The answer is YES. Notice that $\frac{7}{4} = 1.75$ seems to be higher than the terms after it sequence of ratios. After that, the ratio stabalizes and oscillates around phi(the golden ratio). $a_1 = 1 < 1.75$ is clearly satisfied. $1.75^2$ also happends to be slighly larger than 3, so the case $n=2$ is satisfied.

People who came up with this fraction just picked an upper bound on the ratios so that students don't run into problems using induction. It would be okay to pick a fraction such as $\frac{18}{11}$, rule out the case $a_2=3 > 1.637^2 = 2.67..$ , and proceed with induction to show that $a_n \lt {(\frac{18}{11})}^{n}$ for $n \geq 3$. Choosing this number, as well any number after it, doesn't cover some early cases, so it's less desirable.

But you may ask "why not pick 3?" 3 works indeed, but that would make the induction too easy as well as make the bound on growth of Lucas Sequence too weak.

So there you have $\frac{7}{4}$. I hope I answered both of your questions :)