Page 120 - MATINF Nr. 13-14

120 M.N. Popescu

Cum
ln (n) ln (n + 1) − ln (n) n + 1
lim = lim = lim ln = ln (1) = 0,
n→∞ n n→∞ (n + 1) − n n→∞ n

a
avem c˘ ∃n 0 ≥ 2 astfel ca
ln (n!)
−1 + ln (n) ≤ ≤ ln (n) + 1, ∀n ≥ n 0 .
n

Din cele anterioare

X ln (p)
−1 + ln (n) − C < < ln (n) + 1 + ln (4) , ∀n ≥ n 0 ,
p
p≤n
p prim

deci
ln (p)
X
= ln (n) + O (1) .
p n→∞
p≤n
p prim

20. a. Folosim doar ideea de la punctul 16. Fie

2 = p 0 < p 1 < . . . < p l ≤ n

numerele prime din {2, . . . , n}. Avem

n p k+1 n
Z l−1 Z Z
R (t) X R (t) R (t)
dt = dt + dt
t(ln (t)) 2 t(ln (t)) 2 t(ln (t)) 2
2 k=0 p k p l
p k+1 ! n !
l−1 Z k Z l
X 1 X ln (p j ) 1 X ln (p j )
= 2 − ln (t) dt + 2 − ln (t) dt
t(ln (t)) p j t(ln (t)) p j
k=0 j=0 j=0
p k p l
 
! p k+1 ! n n

l−1 k Z l Z Z
X X ln (p j ) 1 X ln (p j ) 1 1
=  2 dt + 2 dt − dt

p j t(ln (t)) p j t(ln (t)) t ln (t)
k=0 j=0 j=0
p k p l 2
l−1 " k ! Ç å# l ! Ç å
n
X X ln (p j ) −1 p k+1 X ln (p j ) −1 n
= + − ln (ln (t))
p j ln (t) p j ln (t) 2
k=0 j=0 p k j=0 p l
l−1 " k k #
X 1 X ln (p j ) 1 X ln (p j )
= − −
ln (p k+1 ) p j ln (p k ) p j
k=0 j=0 j=0
l l
1 X ln (p j ) 1 X ln (p j )
− + − ln 2 (n) + ln 2 (2)
ln (n) p j ln (p l ) p j
j=0 j=0
" #
l−1 k+1 k l−1
X 1 X ln (p j ) 1 X ln (p j ) X 1
= − − +
ln (p k+1 ) p j ln (p k ) p j p k+1
k=0 j=0 j=0 k=0
l l
1 X ln (p j ) 1 X ln (p j )
− + − ln 2 (n) + ln 2 (2)
ln (n) p j ln (p l ) p j
j=0 j=0