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
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