1.4. The Mertens estimates 13
Proof. Partial summation yields
B
n=1
f(n) = Bf(B) −
B
1
[u]f (u) du
and
A
n=1
f(n) = Af(A) −
A
1
[u]f (u) du.
Then
B
A
[u]f (u) du = uf(u)
B
A
−
B
n=A+1
f(n)
after taking the difference. Now
B
A
u −
1
2
f (u) dt = u −
1
2
f(u)
B
A
−
B
A
f(u) du
by integration by parts. Subtracting the next to last formula from the last
formula yields the Euler-Maclaurin summation formula.
Despite the fact that anything obtainable from the Euler-Maclaurin sum-
mation formula may also be obtained by partial summation, resort to the
former is sometimes more convenient. Moreover, repeated integration by
parts in the Euler-Maclaurin summation formula yields a technique for ob-
taining precise approximations to sums. We make no use of this, so it is not
covered here.
The next two results are due to Mertens. These depend on the Cheby-
shev bound ψ(x) = O(x) in an essential way.
Proposition 1.9. The estimates
p≤x
log(p)
p
=
n≤x
Λ(n)
n
+ O(1) = log(x) + O(1)
hold.
Proof. First
x
n≤x
Λ(n)
n
=
n≤x
Λ(n)
x
n
+
O⎝
⎛
n≤x
Λ(n)⎠
⎞
because 0 ≤ x − [x] 1. Now
T(x) =
m≤x
Λ(m)
x
m
