Abel Summation

Phelpedo

I've heard that Abel Summation is the discrete analogue to integration by parts, but I've never seen it in a form that recalls the relation, and google gives absolutely nothing.

**Posted:** Fri Sep 28, 2007 6:14 pm

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Chuck Norris can divide by 0.

Kent Merryfield

Try this: If $A_n=\sum_{k=1}^na_k,$ then:

$\sum_{k=1}^na_kb_k= A_nb_n+\sum_{k=1}^{n-1}A_k(b_k-b_{k+1})$

There are several minor tweaks you can give to this formula, and the starting index could vary. The analogy to integration by parts is there: $a_n,$ which we sum, is analogous to $dv,$ and the part of $u,$ which we would have differentiated, is taken by $b_n,$ which we difference. We soak up the minus sign intrinsic to integration by parts by writing the difference backwards, as $b_k-b_{k+1}.$

Check that this works for $n=1$ or $n=2.$ If you're trying to reconstruct this from having forgotten it, that's what you should focus on. The hard part is getting the right formula; once you have that, the inductive proof that it is correct writes itself.

**Posted:** Fri Sep 28, 2007 11:11 pm