裂項和

裂项求和（Telescoping sum）是一個非正式的用語，指一種用來計算級數的技巧：每項可以分拆，令上一項和下一項的某部分互相抵消，剩下頭尾的項需要計算，從而求得級數和。 ∑ i = 1 n ( a i − a i + 1 ) = ( a 1 − a 2 ) + ( a 2 − a 3 ) + … + ( a n − a n + 1 ) = a 1 − a n + 1 . {\displaystyle \sum _{i=1}^{n}(a_{i}-a_{i+1})=(a_{1}-a_{2})+(a_{2}-a_{3})+\ldots +(a_{n}-a_{n+1})=a_{1}-a_{n+1}.} 裂項積（Telescoping product）也是差不多的概念： ∏ i = 1 n a i a i + 1 = a 1 a n + 1 {\displaystyle \prod _{i=1}^{n}{\frac {a_{i}}{a_{i+1}}}={\frac {a_{1}}{a_{n+1}}}} 1 a ( a + b ) = 1 b ( 1 a − 1 a + b ) {\displaystyle {\frac {1}{a(a+b)}}={\frac {1}{b}}\left({\frac {1}{a}}-{\frac {1}{a+b}}\right)} a k = 1 a − 1 ( a k + 1 − a k ) {\displaystyle a^{k}={\frac {1}{a-1}}(a^{k+1}-a^{k})} cos ⁡ k x = 1 2 sin ⁡ x 2 [ sin ⁡ ( k + 1 2 ) x − sin ⁡ ( k − 1 2 ) x ] {\displaystyle \cos kx={\frac {1}{2\sin {\frac {x}{2}}}}\left[\sin \left(k+{\frac {1}{2}}\right)x-\sin \left(k-{\frac {1}{2}}\right)x\right]} sin ⁡ k x = 1 2 sin ⁡ x 2 [ cos ⁡ ( k − 1 2 ) x − cos ⁡ ( k + 1 2 ) x ] {\displaystyle \sin kx={\frac {1}{2\sin {\frac {x}{2}}}}\left[\cos \left(k-{\frac {1}{2}}\right)x-\cos \left(k+{\frac {1}{2}}\right)x\right]} （三角恒等式） C k n = C k n − 1 + C k − 1 n − 1 {\displaystyle C_{k}^{n}=C_{k}^{n-1}+C_{k-1}^{n-1}} （帕斯卡法則） 1 C k n = n + 1 n + 2 ( 1 C k n + 1 + 1 C k + 1 n + 1 ) {\displaystyle {\frac {1}{C_{k}^{n}}}={\frac {n+1}{n+2}}\left({\frac {1}{C_{k}^{n+1}}}+{\frac {1}{C_{k+1}^{n+1}}}\right)} 若有 a k = b k − b k + 1 {\displaystyle a_{k}=b_{k}-b_{k+1}} ，则 ∑ k = m n a k = b m − b n + 1 {\displaystyle \sum _{k=m}^{n}a_{k}=b_{m}-b_{n+1}} 1 1 ⋅ 2 + 1 2 ⋅ 3 + 1 3 ⋅ 4 + ⋯ + 1 n ⋅ ( n + 1 ) = ∑ k = 1 n 1 k ( k + 1 ) = ∑ k = 1 n ( 1 k − 1 k + 1 ) = ( 1 1 − 1 2 ) + ( 1 2 − 1 3 ) + ( 1 3 − 1 4 ) + ⋯ + ( 1 n − 1 n + 1 ) = 1 − 1 n + 1 {\displaystyle {\frac {1}{1\cdot 2}}+{\frac {1}{2\cdot 3}}+{\frac {1}{3\cdot 4}}+\cdots +{\frac {1}{n\cdot (n+1)}}=\sum _{k=1}^{n}{\frac {1}{k(k+1)}}=\sum _{k=1}^{n}({\frac {1}{k}}-{\frac {1}{k+1}})=({\frac {1}{1}}-{\frac {1}{2}})+({\frac {1}{2}}-{\frac {1}{3}})+({\frac {1}{3}}-{\frac {1}{4}})+\cdots +({\frac {1}{n}}-{\frac {1}{n+1}})=1-{\frac {1}{n+1}}} 若有 a k = b k + b k + 1 {\displaystyle a_{k}=b_{k}+b_{k+1}} ，则 ∑ k = m n ( − 1 ) k a k = ( − 1 ) m b m + ( − 1 ) n + 1 b n + 1 {\displaystyle \sum _{k=m}^{n}(-1)^{k}a_{k}=(-1)^{m}b_{m}+(-1)^{n+1}b_{n+1}} ∑ k = 1 2 n ( − 1 ) k − 1 C 2 n k = 2 n + 1 2 n + 2 ∑ k = 1 2 n ( − 1 ) k − 1 ( 1 C 2 n + 1 k + 1 C 2 n + 1 k + 1 ) = 2 n + 1 2 n + 2 [ 1 C 2 n + 1 1 + ( − 1 ) 2 n − 1 C 2 n + 1 2 n + 1 ] = 2 n + 1 2 n + 2 ( − 2 n 2 n + 1 ) = − n n + 1 {\displaystyle \sum _{k=1}^{2n}{\frac {(-1)^{k-1}}{C_{2n}^{k}}}={\frac {2n+1}{2n+2}}\sum _{k=1}^{2n}(-1)^{k-1}\left({\frac {1}{C_{2n+1}^{k}}}+{\frac {1}{C_{2n+1}^{k+1}}}\right)={\frac {2n+1}{2n+2}}\left[{\frac {1}{C_{2n+1}^{1}}}+{\frac {(-1)^{2n-1}}{C_{2n+1}^{2n+1}}}\right]={\frac {2n+1}{2n+2}}\left({\frac {-2n}{2n+1}}\right)={\frac {-n}{n+1}}} 0 = ∑ n = 1 ∞ 0 = ∑ n = 1 ∞ ( 1 − 1 ) = 1 + ∑ n = 1 ∞ ( − 1 + 1 ) = 1 {\displaystyle 0=\sum _{n=1}^{\infty }0=\sum _{n=1}^{\infty }(1-1)=1+\sum _{n=1}^{\infty }(-1+1)=1\,} 這是錯誤的。將每項重組的方法只適用於獨立的項趨近0。 防止這種錯誤，可以先求首N項的值，然後取N趨近無限的值。 ∑ n = 1 N 1 n ( n + 1 ) = ∑ n = 1 N 1 n − 1 n + 1 {\displaystyle \sum _{n=1}^{N}{\frac {1}{n(n+1)}}=\sum _{n=1}^{N}{\frac {1}{n}}-{\frac {1}{n+1}}\,} = ( 1 − 1 2 ) + ( 1 2 − 1 3 ) + ⋯ + ( 1 N − 1 N + 1 ) {\displaystyle =\left(1-{\frac {1}{2}}\right)+\left({\frac {1}{2}}-{\frac {1}{3}}\right)+\cdots +\left({\frac {1}{N}}-{\frac {1}{N+1}}\right)\,} = 1 + ( − 1 2 + 1 2 ) + ( − 1 3 + 1 3 ) + ⋯ + ( − 1 N + 1 N ) − 1 N + 1 {\displaystyle =1+\left(-{\frac {1}{2}}+{\frac {1}{2}}\right)+\left(-{\frac {1}{3}}+{\frac {1}{3}}\right)+\cdots +\left(-{\frac {1}{N}}+{\frac {1}{N}}\right)-{\frac {1}{N+1}}\,} = 1 − 1 N + 1 → 1   a s   N → ∞ . {\displaystyle =1-{\frac {1}{N+1}}\to 1\ \mathrm {as} \ N\to \infty .\,} ∑ n = 1 N sin ⁡ ( n ) = ∑ n = 1 N 1 2 csc ⁡ ( 1 2 ) [ 2 sin ⁡ ( 1 2 ) sin ⁡ ( n ) ] {\displaystyle \sum _{n=1}^{N}\sin \left(n\right)=\sum _{n=1}^{N}{\frac {1}{2}}\csc \left({\frac {1}{2}}\right)\left[2\sin \left({\frac {1}{2}}\right)\sin \left(n\right)\right]} = 1 2 csc ⁡ ( 1 2 ) ∑ n = 1 N [ cos ⁡ ( 2 n − 1 2 ) − cos ⁡ ( 2 n + 1 2 ) ] {\displaystyle ={\frac {1}{2}}\csc \left({\frac {1}{2}}\right)\sum _{n=1}^{N}\left[\cos \left({\frac {2n-1}{2}}\right)-\cos \left({\frac {2n+1}{2}}\right)\right]} = 1 2 csc ⁡ ( 1 2 ) [ cos ⁡ ( 1 2 ) − cos ⁡ ( 2 N + 1 2 ) ] {\displaystyle ={\frac {1}{2}}\csc \left({\frac {1}{2}}\right)\left[\cos \left({\frac {1}{2}}\right)-\cos \left({\frac {2N+1}{2}}\right)\right]} 唐秀农. 裂项法求和的一般原理和法则. 数学教学通讯. 2013, (9) [2014-06-17]. （原始内容存档于2014-07-14）.  及万会 张来萍 杨春艳. 封闭形和式初步.  https://web.archive.org/web/20060902104046/http://www.math.wisc.edu/~rhoades/Notes/buFall2005putnam.pdf