題目 2. (1−14)(1−19)(1−116)⋯(1−1n2)⋯=?\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\cdots\left(1-\dfrac{1}{n^2}\right)\cdots=?(1−41)(1−91)(1−161)⋯(1−n21)⋯=? 解答 解法一 思路 展開 先取前 NNN 項部分乘積，將每一項分解為 (n−1)(n+1)/n2(n-1)(n+1)/n^2(n−1)(n+1)/n2，再將相鄰項消去。 答題過程 展開 第 nnn 項可寫成 1−1n2=(n−1)(n+1)n2=n−1n⋅n+1n.1-\frac1{n^2}=\frac{(n-1)(n+1)}{n^2} =\frac{n-1}{n}\cdot\frac{n+1}{n}.1−n21=n2(n−1)(n+1)=nn−1⋅nn+1. 所以前 NNN 項乘積為 ∏n=2N(1−1n2)= (12⋅23⋯N−1N)(32⋅43⋯N+1N)= 12⋅N+1N.\begin{align*} \prod_{n=2}^N\left(1-\frac1{n^2}\right) =&\,\left(\frac12\cdot\frac23\cdots\frac{N-1}{N}\right) \left(\frac32\cdot\frac43\cdots\frac{N+1}{N}\right)\\[4mm] =&\,\frac12\cdot\frac{N+1}{N}. \end{align*}n=2∏N(1−n21)==(21⋅32⋯NN−1)(23⋅34⋯NN+1)21⋅NN+1. 令 N→∞N\to\inftyN→∞，即得無窮乘積為 12.\boxed{\frac12}.21.
