18): x→∞lim X 1≤k≤x µ(k) k logx k

Problem 12203

(American Mathematical Monthly, Vol.127, October 2020) Proposed by R. Tauraso (Italy).

Letm be a nonnegative integer, and let µ be the M¨obius function on Z⁺, defined by setting µ(k) equal to(−1)^rifk is the product of r distinct primes and equal to 0 if k has a square prime factor.

Evaluate

n→∞lim 1 ln^m(n)

n

X

k=1

µ(k)

k ln^m+1n k

.

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

Solution. We will show by induction on m that the given limit is equal to m + 1.

The case when m = 0 is known (see H. Iwaniec Lectures on the Riemann Zeta Function p. 18):

x→∞lim X

1≤k≤x

µ(k) k logx

k

= lim

x→∞log(x) X

1≤k≤x

µ(k) k − lim

x→∞

X

1≤k≤x

µ(k)

k log(k) = 0 − (−1) = 1.

Note that if f is a function defined on [1, +∞) and F (x) :=P

1≤k≤xf ^x_k
then the following M¨obius inversion formula holds

X

1≤k≤x

µ(k)Fx k

= X

1≤k≤x

µ(k) X

1≤j≤x/k

f x/k j



= X

1≤n≤x

fx n

 X

k|n

µ(k) = f (x) (1)

with n = jk. Now assume that m ≥ 1 and let f (x) = x log^m(x) then F (x)

x = X

1≤k≤x

1

klog^mx k

= x X

1≤k≤x

1 k

m

X

j=0

m j



log^m−j(x) log^j(k)(−1)^j

=

m

X

j=0

m j



(−1)^jlog^m−j(x) X

1≤k≤x

log^j(k) k

=

m

X

j=0

m j



(−1)^jlog^m−j(x) log^j+1(x)

j + 1 + cj+ O log^j(x) x

!!

= 1

m + 1

m

X

j=0

m + 1 j + 1



(−1)^jlog^m+1(x) +

m

X

j=0

m j



(−1)^jcjlog^m−j(x) + O log^m(x) x



=log^m+1(x) m + 1 +

m

X

j=0

m j



(−1)^jcjlog^m−j(x) + O log^m(x) x



where we used the fact that for any j ≥ 0 there exists cj ∈R such that for x → +∞,

X

1≤k≤x

log^j(k)

k =log^j+1(x)

j + 1 + cj+ O log^j(x) x

! .

Therefore, by (1), log^m(x) = f (x)

x = 1 x

X

1≤k≤x

µ(k)Fx k



= 1

m + 1 X

1≤k≤x

µ(k)

k log^m+1x k

+

m

X

j=0

m j

 (−1)^jcj

X

1≤k≤x

µ(k)

k log^m−jx k

+ O(1) (2)

because for some constant C 

X

1≤k≤x

µ(k) O (log^m(x/k))      

≤C X

1≤k≤x

log^m(x/k)

≤C log^m(x) + C Z x

1

log^m(x/t) dt

≤C log^m(x) + Cx Z +∞

1

log^m(y)

y² dy = O(x).

SinceP

1≤k≤x µ(k)

k = O(1) and, by the inductive step, for 0 ≤ j ≤ m − 1, X

1≤k≤x

µ(k)

k log^m−jx k

= O(log^m−j−1(x)),

after dividing by log^m(x) both sides of (2) we find

1 = 1

m + 1· 1 log^m(x)

X

1≤k≤x

µ(k)

k log^m+1x k

+

m

X

j=0

m j



(−1)^jcj·o(1) + o(1).

Therefore

x→∞lim 1 log^m(x)

X

1≤k≤x

µ(k)

k log^m+1x k

= m + 1

and the proof is complete.