# Sequences and Series/Series and integration

Theorem (Abelian partial summation):

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

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

Then for ${\displaystyle x\geq 1}$ we have

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

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

{\displaystyle {\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

{\displaystyle {\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

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