A. 1
B. \(\dfrac{1}{2}\)
C. \(\dfrac{1}{4}\)
D. \(\dfrac{3}{2}\)
B
Ta có \(1 - \dfrac{1}{{{k^2}}} = \dfrac{{\left( {k - 1} \right)\left( {k + 1} \right)}}{{{k^2}}}\) nên ta suy ra
\(\begin{array}{l}\left[ {\left( {1 - \dfrac{1}{{{2^2}}}} \right)\left( {1 - \dfrac{1}{{{3^2}}}} \right)...\left( {1 - \dfrac{1}{{{n^2}}}} \right)} \right]\\ = \dfrac{{1.3}}{{{2^2}}}.\dfrac{{2.4}}{{{3^2}}}...\dfrac{{\left( {n - 1} \right)\left( {n + 1} \right)}}{{{n^2}}} = \dfrac{{\left( {n + 1} \right)}}{{2n}}\end{array}\)
\(\lim \left[ {\left( {1 - \dfrac{1}{{{2^2}}}} \right)\left( {1 - \dfrac{1}{{{3^2}}}} \right)...\left( {1 - \dfrac{1}{{{n^2}}}} \right)} \right] = \lim \dfrac{{n + 1}}{{2n}} = \dfrac{1}{2}\)
Câu hỏi trên thuộc đề trắc nghiệm dưới đây !
Copyright © 2021 HOCTAP247