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
