Problem 12114
(American Mathematical Monthly, Vol.126, May 2019) Proposed by Z. Franco (USA).
Letn be a positive integer, and let An = {1/n, 2/n, . . . , n/n}. Let anbe the sum of the numerators inAn when these fractions are expressed in lowest terms. FindP∞
n=1an/n4.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. More generally, we show that for any s > 3
∞
X
n=1
an
ns = ζ(s) 2
1 + ζ(s − 2) ζ(s − 1)
.
We have that
an=
n
X
k=1
k
gcd(n, k) = 1 +1 2
n−1
X
k=1
k gcd(n, k)+
n−1
X
k=1
k gcd(n, n − k)
!
= 1 +1 2
n−1
X
k=1
k gcd(n, k)+
n−1
X
k=1
n − k gcd(n, k)
!
= 1 + 1 2
n−1
X
k=1
n
gcd(n, k) = 1 + bn
2 where
bn=
n
X
k=1
n
gcd(n, k)=X
d|n
nϕ(n/d)
d =X
d|n
dϕ(d)
is multiplicative . Since for any prime p and for any positive integer r,
bpr = 1 +
r
X
k=1
pkϕ(pk) = 1 +
r
X
k=1
(p2k− p2k−1) = p2r+1+ 1 p + 1 ,
it follows that for s > 3, the Dirichlet series of the multiplicative function bn reduces to
∞
X
n=1
bn
ns =Y
p
1 +
∞
X
r=1
bpr
prs
!
=Y
p
1 + p p + 1
∞
X
r=1
1
p(s−2)r + 1 p + 1
∞
X
r=1
1 psr
!
=Y
p
1 + p
p + 1· 1
ps−2− 1 + 1
p + 1 · 1 ps− 1
=Y
p
1 −ps−11
1 − p1s
1 −ps−21
=
ζ(s)ζ(s − 2) ζ(s − 1) . Therefore, we may conclude that
∞
X
n=1
an
ns =1 2
∞
X
n=1
1 ns +
∞
X
n=1
bn
ns
!
=ζ(s) 2
1 +ζ(s − 2) ζ(s − 1)
.
