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: .0 .01 .001 .0002 .00003 .000005 .0000008 .00000013 .000000021 .0000000034 etc. ---------------- .01123595505618... = 1/89 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 s[n] = a s[n-1] + b s[n-2] the characteristic polynomial is f(x) = x^2 - ax - b with the two (assumed distinct) roots __________ __________ a + / a^2 + 4b a - / a^2 + 4b r1 = --------------- r2 = --------------- 2 2 It follows that, for the initial values s[0]=0, s[1]=1, the nth term of the recurring sequence is 1 / n n \ s[n] = ------- ( r1 + r2 ) sqrt(D) \ / where D is the discriminant a^2 - 4b. The tabular summation described above, generalized to the base B, then consists of the terms s[0] s[1] s[2] 1 oo s[n] S = ---- + ---- + ---- + .... = - SUM ----- B B^2 B^3 B n=0 B^n Substituting the previous expression for s[n] and collecting terms, we have 1 / oo n oo n \ S = --------- ( SUM (r1/B) - SUM (r2/B) ) B sqrt(D) \ n=0 n=0 / 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 1 / 1 1 \ S = --------- ( -------- - -------- ) B sqrt(D) \ 1 - r1/B 1 - r2/B / 1 / 1 1 \ = ------- ( ------ - ------ ) sqrt(D) \ B - r1 B - r2 / 1 / r1 - r2 \ = ------- ( ------------ ) sqrt(D) \ (B-r1)(B-r2) / Since r1 - r2 equals sqrt(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 can multiply S by B and write the sum oo s[n] B SB = SUM ----- = ---- n=0 B^n f(B) Now, for the Fibonacci numbers the characteristic polynomial 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 s[n] = 3s[n-1] - 5s[n-2] + 7s[n-3] consists of the values 0,0,1,3,4,4,13,47,... and the characteristic polynomial is f(x) = x^3 - 3x^2 + 5x - 7 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 to the right for each successive number (as illustrated above for the Fibonacci numbers) is S = 1/f(10) = 1/(1000-300+50-7) = 1 / 743 This places the familiar "1/89" aspect of the Fibonacci numbers in perspective, and shows that it's just one special case of of much more general result. This can be seen as a generalization of the simple geometric sum, because if we define the 1st-order recurring sequence by s[0]=1 and s[k] = a s[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 oo a^n B 1 SUM --- = ----- = -------- n=0 B^n B - a 1 - (a/B) So far we have considered only recurrences with the initial values 0,0,..,0,1, but we can also generalize to any set of initial values s0,s1,..,s(N-1) for an Nth-order recurrence. Let's write the general characteristic polynomial as f(x) = x^N - c_1 x^(N-1) - c_2 x^(N-2) - ... - c_N and then, calling this f_0(x), let's define the sequence of polynomials f_j(x) for j=1,2,... as follows f_j(x) + c_(N-j) f_{j+1}(x) = ---------------- x For example, if f(x) = x^3 - ax^2 - bx - c then we have f_0(x) = x^3 - ax^2 - bx - c f_1(x) = x^2 - ax - b f_2(x) = x - a f_3(x) = 1 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 oo s[n] f_1(B) s[0] + f_2(B) s[1] + ... + s[N-1] SUM ----- = B ---------------------------------------- n=0 B^n f(B) 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.

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