裂項和

裂項求和（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。

$\sum _{i=1}^{n}(a_{i}-a_{i+1})=(a_{1}-a_{2})+(a_{2}-a_{3})+\ldots +(a_{n}-a_{n+1})=a_{1}-a_{n+1}.$

裂項積（Telescoping product）也是差不多的概念：

$\prod _{i=1}^{n}{\frac {a_{i}}{a_{i+1}}}={\frac {a_{1}}{a_{n+1}}}$

可以用來裂項求和的數學式

${\frac {1}{a(a+b)}}={\frac {1}{b}}\left({\frac {1}{a}}-{\frac {1}{a+b}}\right)$

$a^{k}={\frac {1}{a-1}}(a^{k+1}-a^{k})$

$\cos kx={\frac {1}{2\sin {\frac {x}{2}}}}\left[\sin \left(k+{\frac {1}{2}}\right)x-\sin \left(k-{\frac {1}{2}}\right)x\right]$

$\sin kx={\frac {1}{2\sin {\frac {x}{2}}}}\left[\cos \left(k-{\frac {1}{2}}\right)x-\cos \left(k+{\frac {1}{2}}\right)x\right]$（三角恆等式）

$C_{k}^{n}=C_{k}^{n-1}+C_{k-1}^{n-1}$（帕斯卡法則）

${\frac {1}{C_{k}^{n}}}={\frac {n+1}{n+2}}\left({\frac {1}{C_{k}^{n+1}}}+{\frac {1}{C_{k+1}^{n+1}}}\right)$

求和類型

一般求和

若有$a_{k}=b_{k}-b_{k+1}$，則$\sum _{k=m}^{n}a_{k}=b_{m}-b_{n+1}$

${\frac {1}{1\cdot 2}}+{\frac {1}{2\cdot 3}}+{\frac {1}{3\cdot 4}}+\cdots +{\frac {1}{n\cdot (n+1)}}=\sum _{k=1}^{n}{\frac {1}{k(k+1)}}=\sum _{k=1}^{n}({\frac {1}{k}}-{\frac {1}{k+1}})=({\frac {1}{1}}-{\frac {1}{2}})+({\frac {1}{2}}-{\frac {1}{3}})+({\frac {1}{3}}-{\frac {1}{4}})+\cdots +({\frac {1}{n}}-{\frac {1}{n+1}})=1-{\frac {1}{n+1}}$

交錯求和

若有$a_{k}=b_{k}+b_{k+1}$，則$\sum _{k=m}^{n}(-1)^{k}a_{k}=(-1)^{m}b_{m}+(-1)^{n+1}b_{n+1}$

$\sum _{k=1}^{2n}{\frac {(-1)^{k-1}}{C_{2n}^{k}}}={\frac {2n+1}{2n+2}}\sum _{k=1}^{2n}(-1)^{k-1}\left({\frac {1}{C_{2n+1}^{k}}}+{\frac {1}{C_{2n+1}^{k+1}}}\right)={\frac {2n+1}{2n+2}}\left[{\frac {1}{C_{2n+1}^{1}}}+{\frac {(-1)^{2n-1}}{C_{2n+1}^{2n+1}}}\right]={\frac {2n+1}{2n+2}}\left({\frac {-2n}{2n+1}}\right)={\frac {-n}{n+1}}$

誤用

$0=\sum _{n=1}^{\infty }0=\sum _{n=1}^{\infty }(1-1)=1+\sum _{n=1}^{\infty }(-1+1)=1\,$

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。

$\sum _{n=1}^{N}{\frac {1}{n(n+1)}}=\sum _{n=1}^{N}{\frac {1}{n}}-{\frac {1}{n+1}}\,$

$=\left(1-{\frac {1}{2}}\right)+\left({\frac {1}{2}}-{\frac {1}{3}}\right)+\cdots +\left({\frac {1}{N}}-{\frac {1}{N+1}}\right)\,$

$=1+\left(-{\frac {1}{2}}+{\frac {1}{2}}\right)+\left(-{\frac {1}{3}}+{\frac {1}{3}}\right)+\cdots +\left(-{\frac {1}{N}}+{\frac {1}{N}}\right)-{\frac {1}{N+1}}\,$

$=1-{\frac {1}{N+1}}\to 1\ \mathrm {as} \ N\to \infty .\,$

例子：三角函數

$\sum _{n=1}^{N}\sin \left(n\right)=\sum _{n=1}^{N}{\frac {1}{2}}\csc \left({\frac {1}{2}}\right)\left[2\sin \left({\frac {1}{2}}\right)\sin \left(n\right)\right]$

$={\frac {1}{2}}\csc \left({\frac {1}{2}}\right)\sum _{n=1}^{N}\left[\cos \left({\frac {2n-1}{2}}\right)-\cos \left({\frac {2n+1}{2}}\right)\right]$

$={\frac {1}{2}}\csc \left({\frac {1}{2}}\right)\left[\cos \left({\frac {1}{2}}\right)-\cos \left({\frac {2N+1}{2}}\right)\right]$

外部連結

• https://web.archive.org/web/20060902104046/http://www.math.wisc.edu/~rhoades/Notes/buFall2005putnam.pdf