Page 126 - MATINF Nr. 13-14

126 M.N. Popescu

√
Pentru n ∈ S ∩ [ x, x] avem

ln (x)
√
0 ≤ ln 2 (x) − ln 2 (n) ≤ ln 2 (x) − ln 2 x = ln √ = ln (2) ,
ln ( x)

deci

|ω (n) − ln 2 (x)| ≥ |ω (n) − ln 2 (n)| − |ln 2 (n) − ln 2 (x)| ≥ |ω (n) − ln 2 (n)| − ln (2) ,

prin urmare, deoarece n ∈ S, avem

|ω (n) − ln 2 (x)| |ω (n) − ln 2 (n)| − ln (2) ln (2)
≥ ≥ (ln 2 (n)) 1/4
p p − p
ln 2 (x) ln 2 (n) ln 2 (n)
x
≥ c(ln 2 (n)) 1/4 ≥ c ln 2 √ 1/4 ,
a
unde c ∈ (0, 1) este o constant˘a. Rezult˘a c˘

2 √ 1/2 |ω (n) − ln 2 (x)| 2
c ln 2 x ≤ .
ln 2 (x)
√
Sum˘am dup˘ n ∈ S ∩ [ x, x] s , i obt , inem
a
√ X |ω (n) − ln 2 (x)| 2 X |ω (n) − ln 2 (x)| 2
2 √ 1/2
c ln 2 x · card S ∩ x, x ≤ ≤ ,
√ ln 2 (x) ln 2 (x)
n∈S∩[ x,x] n≤x

deci
!
1 √ 1 X |ω (n) − ln 2 (x)| 2 1
card S ∩ x, x ≤ √ .
x x ln 2 (x) c (ln 2 ( x)) 1/2
2
n≤x
ˆ a
In aceast˘a inegalitate trecem la limit˘ x → ∞ s , i, cu punctul 22.d, obt , inem
1 √
lim card S ∩ x, x = 0.
x→∞ x

Demonstrat , ia este complet˘a.