be a sequence of positive real numbers

Problem 12124

(American Mathematical Monthly, Vol.126, June-July 2019) Proposed by M. Omarjee (France).

Let p >1, and let a1, a2, . . . be a sequence of positive real numbers. Prove that if

n

X

k=1

1 1 + a^p_k = O

 1

1 + a^pn

 ,

then n

X

k=1

1 1 + ak

= O

 1

(1 + a^pn)^1/p

 .

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Solution. We will prove a more general result.

Let p >0 and let (xn)n,(yn)n be two sequences of positive real numbers, such that y_n^p ≤xn for all n ≥1. Then

n

X

k=1

xk = O (xn) =⇒

n

X

k=1

yk = O x^1/p_n 

.

If p > 1, xn = _1+a¹p

n and yn = _1+a¹

n, then y^p_n ≤xn if and only if 1 + a^p_n ≤(1 + an)^p which holds and we are done.

By definition,Pn

k=1xk = O (xn) implies that there is C > 1 such that for n ≥ 1, Sn :=

n

X

k=1

xk < Cxn= C(Sn−Sn−1) =⇒ Sn−1< qSn where q = 1 − 1

C ∈(0, 1).

Hence , for 1 ≤ k ≤ n,

y^p_k≤xk ≤Sk≤qSk+1≤ · · · ≤q^n−kSn< q^n−kCxn =⇒ yk<(q^n−kCxn)^1/p. Finally

n

X

k=1

yk<(Cxn)^1/p

n

X

k=1

q^(n−k)/p<(Cxn)^1/p

∞

X

k=0

q^k/p= C^1/p 1 − q^1/px^1/p_n that isPn

k=1yk = O x^1/pn

.