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。

\sum_{i = 1}^{n} (a_{i} - a_{i + 1}) = (a_{1} - a_{2}) + (a_{2} - a_{3}) + \dots + (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}}

可以用來裂项求和的數學式

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

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

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

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

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

$\frac{1}{C_{k}^{n}} = \frac{n + 1}{n + 2} (\frac{1}{C_{k}^{n + 1}} + \frac{1}{C_{k + 1}^{n + 1}})$ ^[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} + \dots + \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}) + \dots + (\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}^{2 n} \frac{(- 1)^{k - 1}}{C_{2 n}^{k}} = \frac{2 n + 1}{2 n + 2} \sum_{k = 1}^{2 n} (- 1)^{k - 1} (\frac{1}{C_{2 n + 1}^{k}} + \frac{1}{C_{2 n + 1}^{k + 1}}) = \frac{2 n + 1}{2 n + 2} [\frac{1}{C_{2 n + 1}^{1}} + \frac{(- 1)^{2 n - 1}}{C_{2 n + 1}^{2 n + 1}}] = \frac{2 n + 1}{2 n + 2} (\frac{- 2 n}{2 n + 1}) = \frac{- n}{n + 1}

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}

= (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{N} - \frac{1}{N + 1})

= 1 + (- \frac{1}{2} + \frac{1}{2}) + (- \frac{1}{3} + \frac{1}{3}) + \dots + (- \frac{1}{N} + \frac{1}{N}) - \frac{1}{N + 1}

= 1 - \frac{1}{N + 1} \to 1 a s N \to \infty .

\sum_{n = 1}^{N} \sin (n) = \sum_{n = 1}^{N} \frac{1}{2} \csc (\frac{1}{2}) [2 \sin (\frac{1}{2}) \sin (n)]

= \frac{1}{2} \csc (\frac{1}{2}) \sum_{n = 1}^{N} [\cos (\frac{2 n - 1}{2}) - \cos (\frac{2 n + 1}{2})]

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