阿貝爾變換，也叫分部求和法，把兩組數列乘積的和表達成一組數列和與一組數列差。
∑ i = 1 n a i b i = a 1 ( b 1 − b 2 ) + ( a 1 + a 2 ) ( b 2 − b 3 ) + … + ( a 1 + a 2 + ⋯ + a n − 1 ) ( b n − 1 − b n ) + ( a 1 + a 2 + ⋯ + a n ) b n {\displaystyle {\begin{aligned}\sum _{i=1}^{n}a_{i}b_{i}&=a_{1}(b_{1}-b_{2})+(a_{1}+a_{2})(b_{2}-b_{3})+\dots \\~&+(a_{1}+a_{2}+\dots +a_{n-1})(b_{n-1}-b_{n})+(a_{1}+a_{2}+\dots +a_{n})b_{n}\end{aligned}}}
∑ k = 1 n k q k − 1 = − ( 1 ) − ( 1 + q ) − ⋯ − ( 1 + q + ⋯ + q n − 1 ) + n ( 1 + q + ⋯ + q n ) = − 1 − q 1 − q − 1 − q 2 1 − q − ⋯ − 1 − q n 1 − q + n 1 − q n + 1 1 − q = − n − 1 − q − q 2 − ⋯ − q n 1 − q + n 1 − q n + 1 1 − q = 1 − q n + 1 ( 1 − q ) 2 − n q n + 1 1 − q = 1 − ( n + 1 ) q n + 1 + n q n + 2 ( 1 − q ) 2 {\displaystyle {\begin{aligned}\sum _{k=1}^{n}kq^{k-1}&=-(1)-(1+q)-\dots -(1+q+\dots +q^{n-1})+n(1+q+\dots +q^{n})\\~&=-{\frac {1-q}{1-q}}-{\frac {1-q^{2}}{1-q}}-\dots -{\frac {1-q^{n}}{1-q}}+n{\frac {1-q^{n+1}}{1-q}}\\~&=-{\frac {n-1-q-q^{2}-\dots -q^{n}}{1-q}}+n{\frac {1-q^{n+1}}{1-q}}\\~&={\frac {1-q^{n+1}}{(1-q)^{2}}}-{\frac {nq^{n+1}}{1-q}}={\frac {1-(n+1)q^{n+1}+nq^{n+2}}{(1-q)^{2}}}\end{aligned}}}
