Problem 11821
(American Mathematical Monthly, Vol.122, February 2015) Proposed by F. Holland and C. Koester (Ireland).
Let p be a positive integer. Prove that
lim
n→∞
1 2nnp
n
X
k=0
(n − 2k)2pn k
=
p
Y
j=1
(2j − 1).
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 1 2n
n
X
k=0
(n − 2k)2pn k
= 1 2n
2p
X
j=0
2p j
n2p−j·
n
X
k=0
n k
(−2k)j
= 1 2n
2p
X
j=0
2p j
x2p−j (2p − j)!
enx· xj j!
(1 + e−2x)n
= x2p (2p)!
ex+ e−x 2
n
= D2p((cosh(x))n)(0).
The Fa´a di Bruno’s formula says that
D2p(f ◦ g)(x) =X (2p)!
Q2p
j=1(kj)!(j!)kj · (Dkf )(g(x)) ·
2p
Y
j=1
((Djg)(x))kj
where k =P2p
j=1kj and the sum is taken over all 2p-tuples of nonnegative integers (k1, . . . , k2p) for whichP2p
j=1jkj= 2p.
In our case f (x) = xn and g(x) = cosh(x) and since
Dk(xn)(cosh(0)) =
k−1
Y
j=0
(n − j) and Dj(cosh)(0) =
(1 if j is even 0 if j is odd , it follows that we can take k1= k3= · · · = k2p−1= 0.
Hence k =Pp
j=1k2j, the costraint becomesPp
j=1jk2j= p, and
D2p((cosh(x))n)(0) =X (2p)!
Qp
j=1(k2j)!((2j)!)k2j ·
k−1
Y
j=0
(n − j).
The RHS is a polynomial in n of degree the maximum value of k which is equal to p. It is attained only when k2= p and k4= · · · = k2p= 0, therefore we finally find that
n→∞lim 1 2nnp
n
X
k=0
(n − 2k)2pn k
= lim
n→∞
D2p((cosh(x))n)(0)
np = (2p)!
p!(2!)p =
p
Y
j=1
(2j − 1).