X n=1 1 an+1− an

Problem 12004

(American Mathematical Monthly, Vol.124, October 2017) Proposed by M. Omarjee (France).

Let{aⁿ}n≥1 be a strictly increasing sequence of real numbers such that aⁿ≤ n²ln(n) for all n ≥ 1.

Prove that the series

∞

X

n=1

1 an+1− aⁿ diverges.

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

Solution. Assume on the contrary that the series (with positive terms) is convergent.

Then (note that a1≤ 1²ln(1) ≤ 0)

∞

X

n=1

1 an+1− an

=

∞

X

k=1





2^k−1

X

n=2^k−1

1 an+1− an



<+∞ =

∞

X

k=1

1 4(ln(2)k −^a₄¹k)

which implies that there exists a positive integer k such that

2^k−1

X

n=2^k−1

1

an+1− aⁿ < 1 4(ln(2)k −^a₄^k1). Moreover, by Cauchy-Schwarz inequality,





2^k−1

X

n=2^k−1

1 an+1− aⁿ



·





2^k−1

X

n=2^k−1

(an+1− aⁿ)



≥





2^k−1

X

n=2^k−1

1





2

= (2^k− 2^k−1)²= 4^k−1.

Therefore 1

4(ln(2)k −^a₄k¹)>

2^k−1

X

n=2^k−1

1

an+1− aⁿ ≥ 4^k−1 P2^k−1

n=2^k−1(an+1− aⁿ)= 4^k−1 a2^k− a2^k−1

≥ 4^k−1 a2^k− a1

,

that is

a2^k− a1>4^k

ln(2)k −a1

4^k

= 4^kln(2^k) − a1 =⇒ a2^k >(2^k)²ln(2^k)

which contradicts the given inequality an≤ n²ln(n) for n = 2^k.