|
|
$\mathop {\lim }\limits_{n \to \infty } \left( {\prod\limits_{k = 1}^n {\sqrt[{{2^k}}]{2}} } \right)=\mathop {\lim }\limits_{n \to \infty } \left( {\prod\limits_{k = 1}^n {2^{\frac{1}{2^k}}} } \right)=\mathop {\lim }\limits_{n \to \infty } \left( { {2^{\sum_{k=1}^{n}\frac{1}{2^k}}} } \right)= { {2^{\mathop {\lim }\limits_{n \to \infty }\sum_{k=1}^{n}\frac{1}{2^k}}} }$
Ta có
$\sum_{k=1}^{n}\dfrac{1}{2^k}=\dfrac{1}{2}.\dfrac{1-\dfrac{1}{2^n}}{1-\dfrac{1}{2}}=1-\dfrac{1}{2^n}\Rightarrow {\mathop {\lim }\limits_{n \to \infty }\sum_{k=1}^{n}\frac{1}{2^k}}=1 $
Vậy
$\mathop {\lim }\limits_{n \to \infty } \left( {\prod\limits_{k = 1}^n {\sqrt[{{2^k}}]{2}} } \right)=2^1=2$
|