Irrationality of Quadratic Sums Consider a sequence of unit fractions with integer denominators greater than 1. If each denominator is greater than or equal to the square of the preceding denominator, then we will say the sequence is "quadratic".  For example, the sequence 1/3, 1/9, 1/81, 1/6561, 1/43046721, etc., is quadratic. Is the sum of and infinite quadratic sequence necessarily irrational? Obviously if we use the minimal denominators in each case, then as Matthew Hudelsen pointed out, we have 1/d, 1/d2, 1/d4, 1/d8, 1/d16, 1/d32, and so on, which in the base d has the representation This never repeats, so it is not rational. However, if we allow each denominator to exceed the square of the previous denominator, then the proof of irrationality becomes a little more difficult. Dean Hickerson emailed a nice proof of the general question. He notes that the sum of any infinite sequence of unit fractions whose denominators increase rapidly enough that each "tail sum" 1/d­k + 1/dk+1 + 1/dk+2 + ...  for sufficiently large k is strictly less than the corresponding geometric sum 1/dk + (1/dk)2 + (1/dk)3 + ...  then the overall sum is irrational (and this condition is clearly met by my "quadratic" unit fraction sequences.) To prove this irrationality criterion, Hickerson observes that if the infinite sum 1/d1 + 1/d2 + ... is rational, then so are each of the "tails". Thus for every k we have coprime integers Nk and Dk representing the numerator and denominator of the kth tail sum Now suppose that, for all k greater than some constant k0, the kth tail sum is strictly less than the geometric sum of (1/dk)j for j = 1, 2, ...  In other words, suppose This gives the inequality Furthermore, we have so it's clear that the fraction Nk+1/Dk+1 can be written with the integer numerator Nk dk – Dk, which implies that the reduced numerator Nk+1 (after removing any possible common factors with the denominator Dk dk) is less than or equal to this integer. Combining this with our previous (strict) inequality on Nk gives which implies that the numerators of successive "tails" (for all k greater than k0) constitute a strictly decreasing infinite sequence of positive integers beginning with the (presumed) finite integer Nk0, which is impossible. Incidentally, this question relates to the use of the greedy algorithm to expand a given fraction into a sum of unit fractions.  The greediness implies that the sum of all the terms beyond 1/dk must be less than (or equal to) Therefore, dk+1 must be greater than or equal to (dk – 1)dk. Notice that this doesn't quite satisfy the condition of my original question, because we can't say that dk+1 is greater than or equal to dk2, nor does it satisfy the condition of Hickerson's irrationality criterion, so we can't use this criterion to rule out the possibility of an infinite greedy expansion equaling a rational number.  Of course, in this case we can proceed much more simply, noting that the greedy algorithm yields a sequence of remainders with strictly decreasing numerators (since at each stage we subtract 1/d from the current remainder N/D to give the new remainder (Nd – D)/Dd where d is the smallest integer such that Nd – D is positive, and therefore the numerator of the new remainder is essentially d modulo N.) From this we might be tempted to conclude that no infinite sequence of unit fractions with denominators such that dk+1 is equal to or greater than dk2 – dk can have a rational sum, because such a sequence would contradict the fact that the greedy algorithm terminates after a finite number of steps when applied to any rational number. However, although the condition that dk+1 exceeds dk2 – dk is necessary for greediness, it is not quite sufficient, because we have the infinite sequences of unit fractions with denominators given by Sylvester's recurrence dk+1 = dk2 – dk + 1, which have the rational sum 1/(d1 – 1). These sequences are, in a sense, the boundary between greediness and non-greediness, i.e., each denominator is exactly 1 greater than is absolutely necessary for greediness. I wonder if the condition that dk+1 exceeds dk2 – dk + 1 is sufficient to ensure greediness. It's also interesting to consider what happens if we restrict all the denominators in our greedy expansion to be odd.  In this case the sum of all the terms following 1/dk must be no greater than Therefore, in the case of odd greedy expansions we only require that dk+1 be greater than or equal to (1/2)dk2 – dk. According to Richard Guy's "Unsolved Problems In Number Theory", it is not known whether the "odd greedy" unit fraction expansion of a rational number is necessarily finite, so we can only say that if an infinite odd greedy expansion of a rational number exists, there must be infinitely many values of k such that dk+1 is less than dk2 and but greater than or equal to (1/2)dk2 – dk.  The question of odd greedy unit fraction expansions is considered in more detail in the article “The Greedy Algorithm for Unit Fractions”. In general, it would be interesting to know the "minimal" polynomial f() such that the requirement for dk+1 to be at least f(dk) is sufficient to ensure irrationality.  It seems as if it may be the Sylvester sequence plus 1. Return to MathPages Main Menu