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
