# Proof

(related to Proposition: Abelian Partial Summation Method)

• By hypothesis, $a_1,\ldots,a_n$ and $b_1,\ldots,b_n,b_{n+1}$ are given elements of a unit ring $(R,+,\cdot).$
• By setting $A_0:=0$ and $A_k:=\sum_{i=1}^k a_i,$ we get $a_k=A_k-A_{k-1}$ for $k=1,\ldots,n.$
• Thus, it follows
• $\sum_{k=1}^n a_kb_k=\sum_{k=1}^n(A_k-A_{k-1})b_k$ by definition of $A_k,$
• $=\sum_{k=1}^n A_kb_k-\sum_{k=1}^n A_{k-1}b_k$ by distributive law for sums,
• $=\sum_{k=1}^n A_kb_k-\sum_{k=0}^{n-1} A_{k}b_{k+1}$ by replacing the summation index in the second term,
• $=\sum_{k=1}^n A_kb_k-\sum_{k=1}^{n-1} A_{k}b_{k+1}$ because $A_0=0,$
• $=\sum_{k=1}^n A_kb_k-\sum_{k=1}^{n} A_{k}b_{k+1}+A_nb_{n+1}$ because $0=-A_nb_{n+1}+A_nb_{n+1},$
• $=\sum_{k=1}^n A_k(b_k-b_{k-1})+A_nb_{n+1}$ applying distributive law for sums again.

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### References

#### Bibliography

1. Heuser Harro: "Lehrbuch der Analysis, Teil 1", B.G. Teubner Stuttgart, 1994, 11th Edition