裂项和

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。

\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}.

裂项积（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}}}

可以用来裂项求和的数学式

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

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

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

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

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

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

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

{\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}}

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

\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}}

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

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。

\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 .\,

\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]