Problem 11115
(American Mathematical Monthly, Vol.111, November 2004) Proposed by J. Clark (USA).
LetHn be thenth harmonic number, that is, Hn =Pn k=11
k. LetEn= Hn2−
n
X
k=1
1
kHmax(k,n−k). Find lim
n→∞En.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Let Hn(2)=Pn k=1
1
k2. We will show that
En = 1 2H⌊(2)n
2⌋
and therefore
n→∞lim En =1 2
∞
X
k=1
1 k2 =π2
12. Since E1=12H0(2)= 0 and for n > 0 the difference
1 2H⌊(2)n+1
2 ⌋−1 2H⌊(2)n
2⌋ =
0 if n is even 2/(n + 1)2 if n is odd it suffices to prove that the same holds for the difference En+1− En. Note that
En = Hn2−
⌊n2⌋
X
k=1
Hn−k
k −
n
X
k=⌊n2⌋+1
Hk
k .
Assume first that n is even then ⌊n/2⌋ = ⌊(n + 1)/2⌋ = n/2 and
En+1− En = Hn+12 − Hn2−
n 2
X
k=1
Hn+1−k− Hn−k
k −Hn+1
n + 1
= Hn+1+ Hn
n + 1 −
n 2
X
k=1
1
k(n + 1 − k)−Hn+1
n + 1
= Hn
n + 1− 1 n + 1
n 2
X
k=1
1
n + 1 − k+1 k
= 0.
Assume now that n is odd then ⌊n/2⌋ = (n − 1)/2, ⌊(n + 1)/2⌋ = (n + 1)/2 and
En+1− En = Hn+12 − Hn2−
n−1 2
X
k=1
Hn+1−k− Hn−k
k −2Hn+1
2
n + 1 −Hn+1
n + 1+2Hn+1
2
n + 1
= Hn+1+ Hn
n + 1 − 1 n + 1
n−1 2
X
k=1
1
n + 1 − k +1 k
−Hn+1
n + 1
= Hn
n + 1− 1 n + 1
Hn− 2 n + 1
= 2
(n + 1)2.
