Problem 11874
(American Mathematical Monthly, Vol.122, December 2015) Proposed by C. I. V˘alean (Romania).
Evaluate the limits below,
n→∞lim
n−1
X
k=2
ζ(k)
Γ(n − k), lim
n→∞
n−2
X
k=1
(−1)k−1ζ(n − k) Γ(k) .
Solution proposed by Moubinool Omarjee, Lyc´ee Henri IV, Paris, France, and Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, Italy.
We first note that
∞
X
k=2
(ζ(k) − 1) =
∞
X
k=2
∞
X
n=2
1 nk =
∞
X
n=2
∞
X
k=2
1 nk =
∞
X
n=2
1/n2 1 − 1/n =
∞
X
n=2
1 n(n − 1) =
∞
X
n=2
1 n − 1−1
n
= 1.
Moreover for all z ∈ C,
∞
X
k=0
zk−1 Γ(k) =
∞
X
k=1
zk k! = ez. The Cauchy product of these two absolutely convergent series is
∞
X
n=3 n−1
X
k=2
(ζ(k) − 1)zn−k+1 Γ(n − k)
!
= ez
where the convergence is given by Mertens’ theorem.
Since the general term of the series goes to zero, we have that
n→∞lim
n−1
X
k=2
(ζ(k) − 1)zn−k+1 Γ(n − k) = 0.
Therefore
n→∞lim
n−1
X
k=2
ζ(k)zn−k+1 Γ(n − k) = lim
n→∞
n−1
X
k=2
zn−k+1
Γ(n − k)+ lim
n→∞
n−1
X
k=2
(ζ(k) − 1)zn−k+1 Γ(n − k) = lim
n→∞
n−2
X
k=1
zk−1 Γ(k) = ez. Finally we apply this result for z = 1 and z = −1 to the given limits,
n→∞lim
n−1
X
k=2
ζ(k) Γ(n − k)
z=1= e and lim
n→∞
n−2
X
k=1
(−1)k−1ζ(n − k) Γ(k) = lim
n→∞
n−1
X
k=2
ζ(k)(−1)n−k−1 Γ(n − k)
z=−1= e−1.
