1.3. The Abel summation formula 5

1.3. The Abel summation formula

Proposition 1.4. (Abel summation formula) Let {an}n≥1 be a sequence of

complex numbers and let f : [1, +∞) −→ C. For each real number x ≥ 1,

let

A(x) =

n≤x

an

and assume that f(x) has a continuous derivative for x ≥ 1. Then

(1.4)

n≤x

anf(n) = A(x)f(x) −

x

1

A(t)f (t)dt.

Proof. We first assume that x = N, an integer. Then

n≤N

anf(n) = A(1)f(1) + (A(2) − A(1))f(2)

+ · · · + (A(N) − A(N − 1))f(N)

= A(1)(f(1) − f(2)) + · · · + A(N − 1)(f(N − 1) − f(N))

+A(N)f(N).

Next observe that f(i + 1) − f(i) =

i+1

i

f (t)dt for i = 1,...,N − 1, and

that A(t) is constant on the interval [i, i + 1). Therefore,

n≤N

anf(n) = A(N)f(N) −

N−1

i=1

A(i)

i+1

i

f (t)dt

= A(N)f(N) −

N−1

i=1

i+1

i

A(t)f (t)dt

= A(N)f(N) −

N

1

A(t)f (t)dt.

This proves relation (1.4) when x = N is an integer. So, let us now assume

that x is not an integer. Set N = x . On the other hand, since A(t) is

constant on the interval [N, x], the right-hand side of (1.4) can be written

as

A(x)f(x) −

x

1

A(t)f (t)dt

= A(x)f(x) −

x

N

A(t)f (t)dt −

N

1

A(t)f (t)dt

= A(x)f(x) − A(N)

x

N

f (t)dt −

N

1

A(t)f (t)dt

= A(x)f(x) − A(N)(f(x) − f(N)) −

N

1

A(t)f (t)dt