裂項和

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

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

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

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

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

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

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

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。

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

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

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

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

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

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

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

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