裂项求和(Telescoping sum)是一个非正式的用语,指一种用来计算级数的技巧:每项可以分拆,令上一项和下一项的某部分互相抵消,剩下头尾的项需要计算,从而求得级数和。
![{\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}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a991401ddac529faaedfdee3499cf55861e8f81)
裂项积(Telescoping product)也是差不多的概念:
![{\displaystyle \prod _{i=1}^{n}{\frac {a_{i}}{a_{i+1}}}={\frac {a_{1}}{a_{n+1}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2b2d63e2d125869a89d6218ed8bd751363951139)
可以用来裂项求和的数学式
(三角恒等式)[1]
(帕斯卡法则)
[2]
求和类型
一般求和
若有
,则
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ab5ed240c9c90f85ac53f55e95fd7a293398c1b6)
交错求和
若有
,则
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f183e201651bf2344f3cef2e5864daa4ea41f65a)
误用
![{\displaystyle 0=\sum _{n=1}^{\infty }0=\sum _{n=1}^{\infty }(1-1)=1+\sum _{n=1}^{\infty }(-1+1)=1\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8b655a5b26e1e876c432c83d01f255db89166930)
这是错误的。将每项重组的方法只适用于独立的项趋近0。
防止这种错误,可以先求首N项的值,然后取N趋近无限的值。
![{\displaystyle \sum _{n=1}^{N}{\frac {1}{n(n+1)}}=\sum _{n=1}^{N}{\frac {1}{n}}-{\frac {1}{n+1}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1a27a22f5239836c9fb9bcc2f3c58a5f40c3c319)
![{\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)\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f8cfd81e43f751b0360edb0c508deb1997cd60ec)
![{\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}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/82c4ae0bb059c7cfeb55f2a7bd383586d3e36f65)
![{\displaystyle =1-{\frac {1}{N+1}}\to 1\ \mathrm {as} \ N\to \infty .\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/83a40ec448ecc6bd60c13d605b43aad66a52266e)
![{\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1b63228a85f7325ecce822758c42523535de4ade)
![{\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/196f2a30eec601e9479c8b1f1f471023ccee3392)
![{\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e1b96461f451616b0e214efd065d29c0d57336c5)
参考资料
- ^ 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.
外部链接