Notice that for n ≥ 2, log(n

Problem 11829

(American Mathematical Monthly, Vol.122, March 2015) Proposed by P. Bracken (USA).

Let {an}n≥1 be a monotone decreasing sequence of real numbers that converges to 0. Prove that P∞

n=1an/n < ∞ if and only if an= O(1/ log(n)) 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 {a_n}_n≥1 is a monotone and decreasing to zero then {S_N}_{N ≥1} and {T_N}_{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,

a_Nlog(N ) ≤ a_N

N

X

n=2

(log(n) − log(n − 1)) ≤ a_N

N

X

n=2

1 n − 1 ≤

N

X

n=2

a_n−1

n − 1 = S_{N −1}≤ S which implies that an= O(1/ log(n)). Moreover

T_N =

N

X

n=1

(a_n− an+1) log(n) =

N

X

n=2

a_nlog(n) −

N +1

X

n=2

a_nlog(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 aNlog(N ) ≤ M for N ≥ 2, then

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

(a_n− a_n+1) log(n) + a_Nlog(N ) = T_{N −1}+ a_Nlog(N ) ≤ T + M,

which implies that S < ∞.