Problem 11078
(American Mathematical Monthly, Vol.111, April 2004) Proposed by D. Knuth (USA).
A positive integer is cube-free if it is not divisible by the cube of any integer greater than 1. Let P∗
denote the summation restricted to the cube-free integers.
(a) EvaluateP∗
n−2. (b) Prove that the sumP∗
n odd(−1)(n−1)/2n−1 converges and determine its value.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
(a) Let N be an integer greater than 1 and define the following Dirichlet series for Re(s) > 1:
FN(s) =
∞
X
n=1
fN(n)
ns with fN(n) = [n is a N th-power free integer], GN(s) =
∞
X
n=1
gN(n)
ns = ζ(N s) with gN(n) = [n is a N th-power integer].
The Dirichlet convolution of fN and gN is
(fN ∗gN)(n) =X
d|n
fN(d)gN(n/d) = 1
because there is a unique way to write an integer n as the product of a N th-power free integer by a N th-power integer. Therefore for Re(s) > 1
FN(s)GN(s) =
∞
X
n=1
(fN ∗gN)(n) ns
∞
X
n=1
1
ns = ζ(s) and
FN(s) = ζ(s)
GN(s)= ζ(s) ζ(N s). In particular for s = 2 and N = 3 we obtain
∗
X 1
n2 = F3(2) = ζ(2)
ζ(6)= π2/6
π6/945= 315 2π4. (b) For n ≥ 1 the arithmetical function
χ(n) =
(−1)(n−1)/2 if n is odd,
0 if n is even
is completely multiplicative and
(χ · fN) ∗ (χ · gN)(n) = X
d|n
χ(d)fN(d)χ(n/d)gN(n/d)
= χ(n)X
d|n
fN(d)gN(n/d) = χ(n)(fN ∗gN)(n).
For Re(s) > 1, let F˜N(s) =
∞
X
n=1
χ(n)fN(n) ns , G˜N(s) =
∞
X
n=1
χ(n)gN(n)
ns =
∞
X
n=1
χ(nN) nN s =
∞
X
n=1
χ(n) nN s = X
n odd
(−1)(n−1)/2 nN s
then
H(s) = ˜˜ FN(s) ˜GN(s) =
∞
X
n=1
χ(n)(fN ∗gN)(n)
ns =
∞
X
n=1
χ(n) ns = X
n odd
(−1)(n−1)/2 ns .
Since χ(1)gN(1) = 1 6= 0, then 1/ ˜GN(s) is a Dirichlet series which converges absolutely for Re(s) >
1/N :
1/ ˜GN(s) =
∞
X
n=1
µ(n)χ(n) nN s =
∞
X
n=1
µ(n1/N)χ(n)gN(n)
ns .
Moreover, the series ˜H(s) converges conditionally in the strip 0 < Re(s) ≤ 1 and absolutely for Re(s) > 1, therefore the product of ˜H(s) and 1/ ˜GN(s) is a Dirichlet series which converges not only for Re(s) > 1 but also for s = 1 (this is the analogous of Mertens’ Theorem for Dirichlet series).
Then by Uniqueness Theorem
χ(n)fN(n) = χ(n) ∗ µ(n1/N)χ(n)gN(n) = χ(n)X
dN|n
µ(n)
and the identity
F˜N(s) = ˜H(s) · 1/ ˜GN(s) holds also for s = 1. In particular for N = 3 we obtain
∗
X
n odd
(−1)(n−1)/2
n = ˜F3(1) = P
n odd(−1)(n−1)/2n−1 P
n odd(−1)(n−1)/2n−3 = π/4 π3/32= 8
π2.
