• Non ci sono risultati.

x for x &gt

N/A
N/A
Protected

Academic year: 2021

Condividi "x for x &gt"

Copied!
1
0
0

Testo completo

(1)

Problem 11773

(American Mathematical Monthly, Vol.121, April 2014) Proposed by M. Omarjee (France).

Given a positive real number a0, let an+1= exp (−Pn

k=0ak) for n ≥ 0. For which values of b does P

n=0(an)b converge?

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

We note that

an+1= exp −

n−1

X

k=0

ak

!

e−an= ane−an= f (an)

where f (x) = xe−x. Since, f (0) = 0 and 0 < f (x) < x for x > 0, it follows that the sequence {an}n≥0is positive and strictly decreasing to 0. Moreover

f (x) = x − x2+ x3/2 + o(x3)

and by Theorem 2 in Effective asymptotic for some nonlinear recurrences and almost doubly- exponential sequences, by Ionascu and Stanica, Acta Mathematica Universitatis Comenianae (2004), we have that

an = 1

n−ln(n)

2n2 + o ln(n) n2

 . Thus,P

n=0(an)bconverge iffP

n=11/nb converge, that is for b > 1. 

Riferimenti