(b) More generally, exhibit an infinite sequence of distinct rational exponents s such thatP x∈Q+f (x)−s is rational

Problem 11068

(American Mathematical Monthly, Vol.111, March 2004) Proposed by H. Wilf (USA).

For a rational number x that equals a/b in lowest terms, let f (x) = ab.

(a) Show that

X

x∈Q⁺

1 f (x)² = 5

2, where the sum extends over all positive rationals.

(b) More generally, exhibit an infinite sequence of distinct rational exponents s such thatP

x∈Q⁺f (x)^−s is rational.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

We first note that

F (s) = X

x∈Q⁺

1 f (x)^s =

∞

X

a, b= 1, gcd(a, b) = 1

1 (a · b)^s.

Moreover for s > 1 we have that

ζ(s)² =

∞

X

a=1

1 a^s

!2

=

∞

X

a,b=1

1 (a · b)^s =

∞

X

d=1

∞

X

a, b= 1, gcd(a, b) = d

1 (a · b)^s

=

∞

X

d=1

1 d^2s ·

∞

X

a, b= 1, gcd(a, b) = 1

1

(a · b)^s = ζ(2s) · F (s).

Therefore if s is a positive even number 2n then the sum converges to

F (2n) = ζ(2n)² ζ(4n) =



(−1)ⁿ⁻¹2²ⁿ⁻¹B2nπ²ⁿ (2n)!

² 

(−1)²ⁿ⁻¹2⁴ⁿ⁻¹B4nπ⁴ⁿ (4n)!



= 4n 2n



· B_2n² 2|B4n| ∈ Q

where Bk is the k-th Bernoulli number (it is rational!). Here there are some values of the function F (2n):

F (2) = 5

2, F (4) = 7

6, F (6) = 715

691, F (8) =7293 7234.