Problem 12026
(American Mathematical Monthly, Vol.125, February 2018) Proposed by M. Bataille (France).
Forn ∈ N, let Hn=Pn
k=11/k and Sn =
n
X
k=1
(−1)n−k k
k
X
j=1
Hj
Find lim
n→∞Sn/ ln(n) and lim
n→∞(S2n− S2n−1).
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. It easy to verify by induction thatPk
j=1Hj= k(Hk− 1) + Hk, hence S2n=
2n
X
k=1
(−1)kHk−
2n
X
k=1
(−1)k+
2n
X
k=1
(−1)kHk
k = Hn
2 +
2n
X
k=1
(−1)kHk
k and
S2n−1= −
2n−1
X
k=1
(−1)kHk−
2n−1
X
k=1
(−1)k−
2n−1
X
k=1
(−1)kHk
k = H2n−Hn
2 − 1 −
2n−1
X
k=1
(−1)kHk
k . Therefore
S2n− S2n−1= Hn+ 1 − H2n−H2n
2n + 2
2n
X
k=1
(−1)kHk
k . Now, as N goes to infinity,
N
X
k=1
(−1)kHk
k =
N
X
k=1
(−1)k k2 +
N
X
k=2
(−1)kHk−1
k
=
N
X
k=1
(−1)k k2 +1
2
N
X
k=2 k−1
X
j=1
(−1)k k
1 j + 1
k − j
=
N
X
k=1
(−1)k k2 +1
2
N −1
X
j=1 N
X
k=j+1
(−1)k j(k − j)
=
N
X
k=1
(−1)k k2 +1
2
N −1
X
j=1
(−1)j j
N −j
X
k′=1
(−1)k′ k′
=
N
X
k=1
(−1)k k2 +1
2
N −1
X
j=1
(−1)j j
2
−1 2
N −1
X
j=1
(−1)j j
N −1
X
k′=N −j+1
(−1)k′
k′ → −π2
12+ln2(2) 2
where
N −1
X
j=1
(−1)j j
N −1
X
k′=N −j+1
(−1)k′ k′
≤ 1
√N → 0.
Hence, since Hn = ln(n) + γ + o(1),
n→∞lim(S2n−S2n−1) = ln(n)+1−ln(2n)−π2
6 +ln2(2)+o(1) → ln2(2)−ln(2)+1−π2
6 ≈ −0.85760324596 and
n→∞lim Sn
ln(n) = lim
n→∞
Hn
2 ln(n) =1 2.
