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