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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