Abel's Summation Formula
Jump to navigation
Jump to search
Theorem
Let $\sequence {a_n}_{n \in \N_{>0} }$ be a sequence in $\R$.
Let $f : \R_{\ge 1} \to \R$ be a continuously differentiable function.
Let $A : \R_{\ge 1} \to \R$ be defined as:
- $\ds \map A x := \sum_{1 \mathop \le n \mathop \le x} a_n$
Then for all $x \in \R_{\ge 1}$:
- $\ds \sum_{1 \mathop \le n \mathop \le x} a_n \map f n = \map A x \map f x - \int _1 ^x \map A u \map {f'} u \rd u$
Proof
![]() | This theorem requires a proof. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Source of Name
This entry was named for Niels Henrik Abel.