Giúp em bài này với ạ!!!!Tính $\mathop {\lim }\limits_{n \to \infty } \left( {1 - \frac{1}{2}} \right)\left( {1 - \frac{1}{{{2^2}}}} \right)...\left( {1 - \frac{1}{{{2^n}}}} \right)$
Tính $\mathop {\lim }\limits_{n \to \infty } \left( {1 - \frac{1}{2}} \right)\left( {1 - \frac{1}{{{2^2}}}} \right)...\left( {1 - \frac{1}{{{2^n}}}} \right)$
Em tham khảo:
Đặt ${{a}_{n}}=\left( 1-\frac{1}{2} \right)\left( 1-\frac{1}{{{2}^{2}}} \right)\left( 1-\frac{1}{{{2}^{3}}} \right)...\left( 1-\frac{1}{{{2}^{n}}} \right).$
Ta có: ${{a}_{n}}=\frac{1}{2}.\frac{3}{4}.\frac{7}{8}...\ge \frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}.\frac{7}{8}...\left( 1-\frac{1}{n} \right)=\frac{1}{n}.$
Mà $\lim \frac{1}{n}=0.$
Ta lại có ${{a}_{n}}=\frac{1}{2}.\frac{3}{4}.\frac{7}{8}...\left( 1-\frac{1}{{{2}^{n}}} \right)\le \left( 1-\frac{1}{{{2}^{n}}} \right)\left( 1-\frac{1}{{{2}^{n}}} \right)...\left( 1-\frac{1}{{{2}^{n}}} \right)={{\left( 1-\frac{1}{{{2}^{n}}} \right)}^{n}}.$
Do $1-\frac{1}{{{2}^{n}}}<1,\forall n\in \mathbb{N}^*$ nên $\lim \left( 1-\frac{1}{{{2}^{n}}} \right)^n=0.$
Theo nguyên lý kẹp thì $\lim \left( {{a}_{n}} \right)=0.$