Problem 12084
(American Mathematical Monthly, Vol.126, January 2019) Proposed by G. Stoica (Canada).
Let{an}n≥1 be a sequence of nonnegative numbers. Prove that{n1Pn
k=1ak}n≥1 is unbounded if and only if there exists a decreasing sequence{bn}n≥1 such thatlimn→∞bn= 0,P∞
n=1bn is finite, andP∞
n=1anbn is infinite. Is the word “decreasing” essential?
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Let An=Pn
k=1ak, then
N
X
n=1
anbn =
N
X
n=1
Anbn−
N
X
n=1
An−1bn =
N−1
X
n=1
An· (bn− bn+1) + ANbN. (1)
Proof of ⇐. Assume that {n1Pn
k=1ak}n≥1 is bounded, i.e. 0 ≤ An ≤ M n for some M ∈ R. Let {bn}n≥1 be any decreasing nonnegative sequence, then by (1),
N
X
n=1
anbn≤ M
N−1
X
n=1
n(bn− bn+1) + N bN
!
= M
N
X
n=1
bn
So ifP∞
n=1bn is finite thenP∞
n=1anbn is finite too and we have a contradiction.
Proof of ⇒. Assume that {1nPn
k=1ak}n≥1is unbounded. Then there is a strictly increasing sequence {nk}k≥1 of positive integers such that Ank≥ knk. Let
bi:=
∞
X
k=j
1
nkk2 for i = nj−1+ 1, . . . , nj with j ≥ 1 where n0= 0. Then {bn}n≥1 is a decreasing nonnegative sequence such thatP∞
n=1bn is finite:
∞
X
n=1
bn=
∞
X
j=1
(nj− nj−1)
∞
X
k=j
1 nkk2 =
∞
X
k=1
1 nkk2
k
X
j=1
(nj− nj−1) =
∞
X
i=1
1 k2 = π2
6 , and, by (1),P∞
n=1anbn is infinite
∞
X
n=1
anbn ≥
∞
X
n=1
An· (bn− bn+1) ≥
∞
X
k=1
knk· 1 nkk2 =
∞
X
k=1
1
k = +∞.
Yes, the word “decreasing” is essential: let
an =
(k if n = 2k with k ∈ N,
0 otherwise, and bn =
(1
k2 if n = 2k with k ∈ N, 0 otherwise.
then {bn}n≥1 is a nonnegative sequence (which is not decreasing) such that limn→∞bn = 0, P∞
n=1bn=P∞ k=1
1
k2 is finite, andP∞
n=1anbn=P∞ k=1
1
k is infinite, but {n1Pn
k=1ak}n≥1is bounded
0 ≤ 1 n
n
X
k=1
ak = 1 n
⌊log2(n)⌋
X
k=1
k≤ log2(n)(log2(n) + 1) 2n <1.
