ΘρϵηΠατπ

Partial Summation (or Abel's Summation)
Let {an} be a sequence of complex numbers and f(t) be a continuously differentiable function. Let A(t)=∑n≤tan, then ∑n≤xanf(n)=A(x)f(x)−∫1xA(t)f′(x)dt
Use the fact stated below, ∑n≤xanf(n)=∑n≤x(A(n)−A(n−1))f(n)=∑n≤xA(n)f(n)−∑n≤x−1A(n)f(n+1)=∑n≤x−1A(n)(f(n)−f(n+1))+A(x)f(x)=A(x)f(x)+∑n≤x−1A(n)∫n+1nf′(t)dt=A(x)f(x)+∑n≤x−1∫n+1nA(n)f′(t)dt=A(x)f(x)−∫1xA(n)f′(t)dt

Remark:

  1. Here an is an Arithmetic function
  2. Fact: an=A(n)−A(n−1)
  3. When x∉ℕ, then A(x)f(x)−f(⌊x⌋)−∫⌊x⌋xA(t)f′(t)dt=0

More General Version
Let 0≤λ1<λ2<... be an increasing real sequence and Cn be a complex sequence. Let C(t)=∑λn≤tCn and f(t) be a continuous derivative, then for x≥λ1, ∑λn≤xCnf(λn)=C(x)f(x)−∫λ1xC(t)f′(x)dt and if C(x)f(x)→0 as x→∞, then ∑n=1∞Cnf(λn)=−∫∞λ1C(t)f′(t)dt