# Sequences and Series/Series and integration

Theorem (Abelian partial summation):

Let $(a_{n})_{n\in \mathbb {N} }$ be a sequence of complex numbers, and let $f:[1,\infty )\to \mathbb {C}$ be differentiable on $(1,\infty )$ . Finally define

$A(x):=\sum _{1\leq n\leq x}a_{n}$ .

Then for $x\geq 1$ we have

$\sum _{1\leq n\leq x}a_{n}f(n)=A(x)f(x)-\int _{1}^{x}A(t)f'(t)dt$ .

Proof: If $m=\lfloor x\rfloor$ , we have

{\begin{aligned}\int _{1}^{x}A(t)f'(t)dt&=\sum _{k=2}^{m}\int _{k-1}^{k}A(t)f'(t)dt+\int _{m}^{x}A(t)f'(t)dt\\&=\sum _{k=2}^{m}\sum _{1\leq n\leq k-1}a_{n}\int _{k-1}^{k}f'(t)dt+\sum _{1\leq n\leq m}a_{n}\int _{m}^{x}f'(t)dt\\&=\sum _{k=2}^{m}A(k-1)(f(k)-f(k-1))+A(x)f(x)-A(m)f(m).\end{aligned}} But

{\begin{aligned}\sum _{k=2}^{m}A(k-1)(f(k)-f(k-1))&=\sum _{k=2}^{m}A(k-1)f(k)-\sum _{k=2}^{m}A(k-1)f(k-1)\\&=\sum _{k=2}^{m}A(k-1)f(k)-\sum _{k=1}^{m-1}A(k)f(k)\\&=\sum _{k=2}^{m-1}f(k)(\underbrace {A(k-1)-A(k)} _{=-a_{k}})+A(m-1)f(m)-A(1)f(1).\end{aligned}} so that

$\int _{1}^{x}A(t)f'(t)dt=A(x)f(x)-\sum _{k=1}^{m}a_{k}f(k)$ . $\Box$ 