Page 104 - RMGO 2

104 Rezolvarea problemelor din num˘arul anterior

MGO 33. Fie s , irurile (x n ) n≥1 , (y n ) n≥1 s , i (z n ) n≥1 de numere reale pozitive,
x n+1 x 1 + x 2 + . . . + x n
astfel ˆıncˆat lim = 1, y n = s , i z n = max {x 1 , x 2 , . . . , x n },
n→∞ x n n
y n+1 z n+1
pentru orice n ≥ 1. Ar˘atat ,i c˘a lim = 1 s , i lim = 1.
n→∞ y n n→∞ z n
Mihai Florea Dumitrescu, Potcoava

y n+1 s n+1 n
Solut ,ie. Avem = · , unde s n = x 1 + x 2 + . . . + x n , ∀n ≥ 1. S , irul
y n s n n + 1
(s n ) este cresc˘ator, deci exist˘a lim s n = l, finit˘a sau +∞. Dac˘a limita l este finit˘a,
n→∞
s n+1 l
atunci l ≥ s 1 = x 1 > 0, deci lim = = 1. Dac˘a l = +∞, atunci aplicˆand
n→∞ s n l
s n+1 s n+2 − s n+1 x n+1
Lema Stolz-Cesaro avem lim = lim = lim = 1. Prin
n→∞ s n n→∞ s n+1 − s n n→∞ x n
y n+1 s n+1 n
urmare, ˆın ambele cazuri avem lim = lim · lim = 1 · 1 = 1.
n→∞ y n n→∞ s n n→∞ n + 1
Evident, z n+1 = max{z n , x n+1 }, ∀n ≥ 1. Dac˘a z n ≥ x n+1 , atunci z n+1 = z n ,
z n+1 z n+1 x n+1 x n+1
deci = 1. Dac˘a z n < x n+1 , atunci z n+1 = x n+1 s , i 1 < = ≤
z n z n z n x n
z n+1 x n+1
(deoarece, evident, z n ≥ x n ). Prin urmare, avem 1 ≤ ≤ max 1, ,
z n x n
z n+1
∀n ≥ 1. Aplicˆand Criteriul cles , telui rezult˘a c˘a lim = 1.
n→∞ z n
Remarc˘am c˘a, ˆın ipotezele date, dac˘a t n = min {x 1 , x 2 , . . . , x n }, pentru orice

x n+1 t n+1
n ≥ 1, atunci, procedˆand ca mai sus, avem min 1, ≤ ≤ 1, ∀n ≥ 1,
x n t n
t n+1
deci s , i lim = 1.
n→∞ t n
MGO 34. Demonstrat ,i c˘a dac˘a matricele A, B, C ∈ M n (Q) verific˘a relat ,ia
2
(B − C) = (A + B)(A + C), atunci det(AB − BA + CA − AC + CB − BC) = 0.
George Mihai, Slatina

2
Solut ,ie. Fie x 1 s , i x 2 r˘ad˘acinile ecuat , iei x −x−1 = 0, deci x 1 +x 2 = 1, x 1 x 2 = −1,
√
1 ± 5
x 1,2 = . Consider˘am matricele U = A + x 1 B + x 2 C s , i V = A + x 2 B + x 1 C.
2
2
2
2
Avem UV = A −B −C +BC +CB +x 1 (BA+AC +BC)+x 2 (AB +CA+CB).
2
2
2
2
2
2
Dar A − B − C + BC + CB = A − (B − C) = A − (A + B)(A + C) =
−(BA+AC +BC), deci UV = (x 1 −1)(BA+AC +BC)+x 2 (AB +CA+CB) =
√
√
x 2 (AB+CA+CB−BA−AC−BC). Evident, det U = a+b 5 s , i det V = a−b 5,
2
2
cu a, b ∈ Q. Pe de o parte, det(UV ) = det U det V = a − 5b ∈ Q, iar pe de alt˘a
n
parte det(UV ) = det (x 2 (AB + CA + CB − BA − AC − BC)) = x det(AB +
2 √
n
CA + CB − BA − AC − BC). Cum x ∈ R \ Q (deoarece x n = c n ± d n 5,
1,2
2
cu c n , d n ∈ Q, iar dac˘a d n = 0 ar rezulta c˘a x 1 = ±x 2 , fals), conchidem c˘a
det(AB + CA + CB − BA − AC − BC) = 0.