Page 113 - MATINF Nr. 13-14

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Rezolvarea problemelor date la concursul de admitere la Ecole Polytechnique, filiera MPI, sesiunea 2024 113

1 1 ln (n)
a
cum < < , ∀n ≥ 3, putem spune c˘ ∃C 2 > 0 s , i n 2 ∈ N astfel ca
n 2 n n
n n
ln (n) X X 1 ln (n)
2
2
−C 2 · + [ln (n) + γ] ≤ ≤ [ln (n) + γ] + C 2 · , ∀n ≥ n 2 . (3)
n ij n
i=1 j=1
1
X
2
Seria este convergent˘a, fie b (= π /6 ) suma ei. Atunci
i 2
i≥1
n Å ã
X 1 X 1 X 1 X 1 1 1
0 < b − = ≤ = − = ,
i 2 i 2 i (i − 1) i − 1 i n
i=1 i≥n+1 i≥n+1 i≥n+1
deci ∃C 3 > 0 s , i n 3 ∈ N astfel ca

n
ln (n) X 1 ln (n)
−C 3 · − b ≤ − ≤ −b + C 3 · , ∀n ≥ n 3 . (4)
n i 2 n
i=1
a
Din (1)-(4) deducem c˘ ∃C 4 > 0 s , i n 4 ∈ N astfel ca
ln (n) 1 X
2
−C 4 · + ln (n) + γ + [ln (n) + γ] − b ≤ ω(σ) 2
n n!
σ∈S n
ln (n)
2
≤ ln (n) + γ + [ln (n) + γ] − b + C 4 · ,
n
2
pentru orice n ≥ n 4 , deci, pentru c = γ + γ − b, avem
1 X 2 2 Å ln (n) ã
ω(σ) = (2γ + 1) ln (n) + c + (ln (n)) + O .
n! n→∞ n
σ∈S n

14. b. Avem

1 X 2 1 X 2 ln (n) X 1 2
(ω (σ) − ln (n)) = ω(σ) − 2 · ω (σ) + · (ln (n)) · n!
n! n! n! n!
σ∈S n σ∈S n σ∈S n
1 X
2
2
= ω(σ) − 2 ln (n) M (X n ) + (ln (n)) ,
n!
σ∈S n
deci, cu punctele 14.a s , i 12,

1 X 2 2
(ω (σ) − ln (n)) = (2γ + 1) ln (n) + c + (ln (n)) − 2 ln (n) [ln (n) + γ]
n! n→∞
σ∈S n
Å ã
2 ln (n)
+ (ln (n)) + O ,
n

deci
1 X 2 Å ln (n) ã
(ω (σ) − ln (n)) = ln (n) + c + O .
n! n→∞ n
σ∈S n