Problem 11865
(American Mathematical Monthly, Vol.122, November 2015) Proposed by Gary H. Chung (USA).
Let{an}n≥1be a monotone decreasing sequence nonnegative real numbers.
Prove thatP∞
n=1an/n < ∞ if and only if limn→∞an= 0 andP∞
n=1(an− an+1) log(n) < ∞.
Solution proposed by Moubinool Omarjee, Lyc´ee Henri IV, Paris, France, and Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, Italy.
Let
SN =
N
X
n=1
an
n and TN =
N
X
n=1
(an− an+1) log(n).
If {an}n≥1is a monotone decreasing sequence nonnegative real numbers then {SN}N ≥1and {TN}N ≥1
are increasing. Let S and T their limits. Notice that for n ≥ 2,
log(n) − log(n − 1) = Z n
n−1
dx x ∈ 1
n, 1 n − 1
.
i) If S < ∞ then for N ≥ 2,
aNlog(N ) ≤ aN
N
X
n=2
(log(n) − log(n − 1)) ≤ aN
N
X
n=2
1 n − 1 ≤
N
X
n=2
an−1
n − 1 = SN −1≤ S which implies that an= O(1/ log(n)) and therefore limn→∞an= 0. Moreover
TN =
N
X
n=1
(an− an+1) log(n) =
N
X
n=2
anlog(n) −
N +1
X
n=2
anlog(n − 1)
=
N
X
n=2
an(log(n) − log(n − 1)) − anlog(n − 1)
≤
N
X
n=2
an n − 1 ≤
N
X
n=2
an−1
n − 1= SN −1≤ S
which implies that T < ∞.
ii) If T < ∞ and limn→∞an= 0 then for k ≥ N ≥ 2,
log(N )(aN− ak+1) = log(N )
k
X
n=N
(an− an+1) ≤
k
X
n=N
log(n)(an− an+1) = Tk− TN −1≤ T
and as k → ∞, we obtain that aNlog(N ) ≤ T . Hence
SN− a1=
N
X
n=2
an n ≤
N
X
n=2
an(log(n) − log(n − 1)) =
N
X
n=2
anlog(n) −
N −1
X
n=2
an+1log(n)
=
N −1
X
n=2
(an− an+1) log(n) + aNlog(N ) = TN −1+ aNlog(N ) ≤ 2T,
which implies that S < ∞.
