An Exploration of the Fibonacci Sequence

By

Kenneth E. Montgomery

Fibonacci sequences are sequences of whole numbers generated by Equation 1:

Equation 1:      , for all

The Fibonacci Sequence is generated by Equation 1, where. A spreadsheet was used to investigate this sequence and the ratios of its terms (Table 1).

 n f(n) f(n+1)/f(n) f(n+2)/f(n) f(n+3)/f(n) 0 1 1 1 1 2 2 2 2 3 3 1.5 3 3 4 5 1.666666667 2.5 5 5 8 1.6 2.666666667 4 6 13 1.625 2.6 4.333333333 7 21 1.615384615 2.625 4.2 8 34 1.619047619 2.615384615 4.25 9 55 1.617647059 2.619047619 4.230769231 10 89 1.618181818 2.617647059 4.238095238 11 144 1.617977528 2.618181818 4.235294118 12 233 1.618055556 2.617977528 4.236363636 13 377 1.618025751 2.618055556 4.235955056 14 610 1.618037135 2.618025751 4.236111111 15 987 1.618032787 2.618037135 4.236051502 16 1597 1.618034448 2.618032787 4.236074271 17 2584 1.618033813 2.618034448 4.236065574 18 4181 1.618034056 2.618033813 4.236068896 19 6765 1.618033963 2.618034056 4.236067627 20 10946 1.618033999 2.618033963 4.236068111 21 17711 1.618033985 2.618033999 4.236067926 22 28657 1.61803399 2.618033985 4.236067997 23 46368 1.618033988 2.61803399 4.23606797 24 75025 1.618033989 2.618033988 4.23606798 25 121393 1.618033989 2.618033989 4.236067976 26 196418 1.618033989 2.618033989 4.236067978 27 317811 1.618033989 2.618033989 4.236067977 28 514229 1.618033989 2.618033989 4.236067978 29 832040 1.618033989 2.618033989 4.236067977 30 1346269 1.618033989 2.618033989 4.236067978

Table 1: The Fibonacci sequence and ratios of its terms

Open the XLS file to explore this problem. In Table 1, we have the first thirty Fibonacci terms and the ratios of sequential terms, every second term and every third term. The values to which these ratios tend to converge are explored.

Careful observation leads to the hypothesis that the ratio of sequential Fibonacci terms approaches the value known as the Golden Ratio and given by:

Which leads us to hypothesize that the ratio of every second term approaches the value:

Thus, we have the ratio of every third term approaching this value, which should be given by:

Setting  and , we have the Lucas Sequence (Table 2).

 n f(n) f(n+1)/f(n) f(n+2)/f(n) f(n+3)/f(n) 0 1 1 3 3 2 4 1.333333 4 3 7 1.75 2.333333 7 4 11 1.571429 2.75 3.666667 5 18 1.636364 2.571429 4.5 6 29 1.611111 2.636364 4.142857 7 47 1.62069 2.611111 4.272727 8 76 1.617021 2.62069 4.222222 9 123 1.618421 2.617021 4.241379 10 199 1.617886 2.618421 4.234043 11 322 1.61809 2.617886 4.236842 12 521 1.618012 2.61809 4.235772 13 843 1.618042 2.618012 4.236181 14 1364 1.618031 2.618042 4.236025 15 2207 1.618035 2.618031 4.236084 16 3571 1.618034 2.618035 4.236062 17 5778 1.618034 2.618034 4.23607 18 9349 1.618034 2.618034 4.236067 19 15127 1.618034 2.618034 4.236068 20 24476 1.618034 2.618034 4.236068 21 39603 1.618034 2.618034 4.236068 22 64079 1.618034 2.618034 4.236068 23 103682 1.618034 2.618034 4.236068 24 167761 1.618034 2.618034 4.236068 25 271443 1.618034 2.618034 4.236068 26 439204 1.618034 2.618034 4.236068 27 710647 1.618034 2.618034 4.236068 28 1149851 1.618034 2.618034 4.236068 29 1860498 1.618034 2.618034 4.236068 30 3010349 1.618034 2.618034 4.236068

Table 2: The Lucas sequence and ratios of its terms

The Lucas sequence can be explored by clicking on the Lucas tab in the Excel spreadsheet. Although the initial values are different, we see that the ratios tend to converge to the same values as those of the Fibonacci Sequence.

Theorem: The ratios of successive terms in a Fibonacci sequence converge to the Golden Ratio. The ratios of pairs of second terms converge to the valueand the ratios of pairs of third terms converge to the value.

Proof:

Part 1.  Proof that the ratio of sequential terms converges to.

Assume

If , then .

Thus, , so

Rearranging, we have:

Solving for L with the quadratic formula, we have:

If , then L<0, which is not possible, so we discard this root.

Thus,

.

Part 2: Proof that the ratios of second terms converge to .

and since,  from part 1:

Part 3: Proof that the ratios of third terms converge to.

Since (from part 2) and since (from part 1), we have:

Thus,