Fibonacci, 1/89, And All That One of the most frequently re-discovered facts about the Fibonacci numbers 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ... is that if we tabulate these numbers in a column, shifting the decimal point one place to the right for each successive number, the sum equals 1/89, as indicated below: This is fairly trivial to prove, but it's an instance of an interesting general pattern involving geometrically weighted sums of the terms of arbitrary linear recurring sequences. In the simple case of Fibonacci numbers, or, a bit more generally, any monic second-order recurrence relation the characteristic polynomial is with the two (assumed distinct) roots It follows that, for the initial values s0 = 0, s1 = 1, the nth term of the recurring sequence is where D is the discriminant A2 – 4B. The tabular summation described above, generalized to the base b, then consists of the terms Substituting the previous expression for sn and collecting terms, we have If the geometric sums converge, (meaning that the roots with the largest magnitude is smaller than the base B), then we can write the sums in closed form Since r1 – r2 equals √D, and noting that the denominator in the parentheses is f(b), because it is the product of terms of the form b minus each root of f, we have Now, as noted above, the characteristic polynomial for the Fibonacci numbers is f(x) = x2 – x – 1, so the sum S for the base 10 is given by setting b = 10, which gives S = 1/f(10) = 1/89. It can be shown that this same summation applies to any (convergent) linear recurring sequence with the initial values 0, 0, ..,0, 1 (assuming the characteristic polynomial has distinct roots). For example, the sequence that satisfies the 3rd order relation consists of the values 0, 0, 1, 3, 4, 4, 13, 47, ..., and the characteristic polynomial is As can easily be verified, the sum of the terms of this sequence, written in base 10, arranged in a column with the decimal point shifted one place to the right for each successive number (as illustrated above for the Fibonacci numbers) is This places the familiar "1/89" aspect of the Fibonacci numbers in perspective, and shows that it's just one special case of much more general result. It can be seen as a generalization of the simple geometric sum, because if we define the 1st-order recurring sequence by s0 = 1 and sk = Ask–1 then we have sn = An, and the characteristic polynomial is f(x) = x – A. So, the general formula reduces to the familiar sum So far we’ve considered only recurrences with the initial values 0, 0, .., 0, 1, but we can also generalize to any set of initial values s0, s1, .., sN–1 for an Nth-order recurrence. Let's write the general characteristic polynomial as and then, calling this f0(x), let's define the sequence of polynomials fj(x) for j = 1, 2, ... as follows For example, if f(x) = x3 – Ax2 – Bx – C then we have In these terms we can write the sum of sn/bn for n = 0 to infinity for any arbitrary linear recurrence (with distinct characteristic roots) and any initial values s0, s1,.., sN–1 as Of course, in our previous considerations we had s0, s1, .., sN–2 all equal to zero, and sN–1 equal to 1, so this reduced to b/f(b). In general we see that the sum equals a linear function of the N initial values, where the kth initial value has a coefficient of b fk+1(b)/f0(b). In the special case of sequences that satisfy the Fibonacci recurrence sn = sn–1 + sn–2 beginning with the arbitrary initial values s0 and s1, the sum is given by [(b–1)s0 + s1] / 89. Return to MathPages Main Menu