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 preceeding denominator, then we will say the
sequence is "quadratic".  For example, the sequence 1/3, 1/9,
 1/81, 1/6561, 1/43046721, ...   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/d^2, 1/d^4, 1/d^8, 
1/d^16, 1/d^32, and so on, which has the representation in the 
base d


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/d[k] + 1/d[k+1] + 1/d[k+2] + ...  for sufficiently large
k is strictly less than the corresponding geometric sum
1/d[k] + (1/d[k])^2 + (1/d[k])^3 + ...  then the overall
sum is irrational (and this condition is clearly met by my
"quadratic" unit fraction sequences.)

To prove this irrationality criterion, Dean observes that if the
infinite sum 1/d[1] + 1/d[2] + ... is rational, then so are each
of the "tails".  Thus for every k we have coprime integers N[k]
and D[k] representing the numerator and denominator of the kth
tail sum
           N[k]       1       1        1
           ----  =  ---- + ------ + ------ + ...
           D[k]     d[k]   d[k+1]   d[k+2]

Now suppose that, for all k greater than some constant k0, the
kth tail sum is strictly less than the geometric sum of (1/d[k])^j
for j=1,2,...  In other words, suppose

   N[k]         1       1        1                  1
   ----   <   ---- + ------ + ------ + ...   =   --------
   D[k]       d[k]   d[k]^2   d[k]^3             d[k] - 1

This gives the inequality

         N[k] d[k]  -  D[k]   <   N[k]

Furthermore, we have

      N[k+1]       N[k]     1        N[k] d[k] - D[k]
      ------   =   ---- - ----   =   ----------------
      D[k+1]       D[k]   d[k]           D[k] d[k]

so it's clear that the fraction N[k+1]/D[k+1] can be written with
the integer numerator N[k] d[k] - D[k], which implies that the
reduced numerator N[k+1] (after removing any possible common
factors with the denominator D[k] d[k]) is less than or equal
to this integer.  Combining this with our previous (strict)
inequality on N[k] gives

        N[k+1]    <=    N[k] d[k] - D[k]    <    N[k]

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
N[k0], 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/d[k]
must be less than (or equal to)

        1/(d[k] - 1) - 1/d[k]   =   1/(d[k](d[k]-1))

Therefore, d[k+1] must be greater than or equal to d[k](d[k]-1). 
Notice that this doesn't quite satisfy the condition of my original 
question, because we can't say that d[k+1] is greater than or equal
to d[k]^2, nor does it satisfy the condition of Dean's irrationality 
criterion, so we can't use this criterion to rule out the possibility 
of an infinite greedy expansion equalling 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 d[k+1] is 
equal to or greater than d[k]^2 - d[k] 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 
d[k+1] exceeds d[k]^2 - d[k] 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 
d[k+1] = d[k]^2 - d[k] + 1, which have the rational sum 1/(d[1]-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 d[k+1] *exceeds* d[k]^2 - d[k] + 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/d[k] must be no greater than

   1/(d[k] - 2) - 1/d[k]   =   2/(d[k](d[k]-2))

Therefore, in the case of odd greedy expansions we only require
that d[k+1] be greater than or equal to (1/2)d[k]^2 - d[k]. 
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 d[k+1] is less than d[k]^2 and but greater than or equal 
to (1/2)d[k]^2 - d[k].  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 d[k+1] to be at 
least f(d[k]) is sufficient to ensure irrationality.  It seems 
as if it may be the Sylvester sequence plus 1.

Return to MathPages Main Menu