Fibonacci, 1/89, And All That 

One of the most frequently rediscovered 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 secondorder recurrence relation 



the characteristic polynomial is 



with the two (assumed distinct) roots 



It follows that, for the initial values s_{0} = 0, s_{1} = 1, the nth term of the recurring sequence is 



where D is the discriminant A^{2} – 4B. The tabular summation described above, generalized to the base b, then consists of the terms 



Substituting the previous expression for s_{n} 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 r_{1} – r_{2} 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) = x^{2} – 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 1storder recurring sequence by s_{0} = 1 and s_{k} = As_{k–1} then we have s_{n} = A^{n}, 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 s_{0}, s_{1}, .., s_{N–1} for an Nthorder recurrence. Let's write the general characteristic polynomial as 



and then, calling this f_{0}(x), let's define the sequence of polynomials f_{j}(x) for j = 1, 2, ... as follows 



For example, if f(x) = x^{3} – Ax^{2} – Bx – C then we have 



In these terms we can write the sum of s_{n}/b^{n} for n = 0 to infinity for any arbitrary linear recurrence (with distinct characteristic roots) and any initial values s_{0}, s_{1},.., s_{N–1} as 



Of course, in our previous considerations we had s_{0}, s_{1}, .., s_{N–2} all equal to zero, and s_{N–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 f_{k+1}(b)/f_{0}(b). In the special case of sequences that satisfy the Fibonacci recurrence s_{n} = s_{n–1} + s_{n–2} beginning with the arbitrary initial values s_{0} and s_{1}, the sum is given by [(b–1)s_{0} + s_{1}] / 89. 
