Page 122 - MATINF Nr. 13-14

122 M.N. Popescu

s , i
∞ ∞ ∞

Z Z Z ∞
R (t) |R (t)| C C
C

dt ≤ dt ≤ dt = − = ,
t(ln (t)) t(ln (t)) t(ln (t))
2 2 2 ln (t) ln (n)
n
n n n
deci
n
Z
C R (t) C
− + c 0 ≤ dt ≤ c 0 + .
ln (n) t(ln (t)) 2 ln (n)
2
20. c. Din punctul 20.a avem
n
1 1 ln (p) Z R (t)
X X
= ln 2 (n) − ln 2 (2) + + dt,
p ln (n) p t(ln (t)) 2
p≤n p≤n 2
p prim p prim
ˆın care (cu punctul 20.b)

n
C Z R (t) C
− + c 0 ≤ dt ≤ c 0 + , ∀n ≥ 2,
ln (n) t(ln (t)) 2 ln (n)
2
¯
s , i (cu punctul 19.d) ∃C ≥ 0 s , i ∃n 0 ≥ 2 astfel ca

¯
¯
C 1 X ln (p) C
− + 1 ≤ ≤ 1 + , ∀n ≥ n 0 .
ln (n) ln (n) p ln (n)
p≤n
p prim
Prin urmare, ∀n ≥ n 0 , avem

¯
¯
C + C X 1 C + C
− + ln 2 (n) − ln 2 (2) + 1 + c 0 ≤ ≤ ln 2 (n) − ln 2 (2) + 1 + c 0 + ,
ln (n) p ln (n)
p≤n
p prim
adic˘
a
Å ã
X 1 1
= ln 2 (n) + c 1 + O ,
p n→∞ ln (n)
p≤n
p prim
unde
c 1 = −ln 2 (2) + 1 + c 0 .

21. a. Deoarece

{n ∈ N ∩ [1, x] : n ≡ 0 (mod q)} = [1, x] ∩ {1q, 2q, . . . , kq, . . .}

s , i
Å ã
x
kq ∈ [1, x] ⇔ k ≤ E ,
q
avem
x
x Å ã x
card {n ∈ N ∩ [1, x] : n ≡ 0 (mod q)} − = E − ,
q q q