Mathematics

If $$f\left ( n \right )= \left [ \left ( n+1 \right )\left ( n+2 \right )...\left ( n+n \right ) \right ]^{1/n}$$, then $$\lim_{n\rightarrow \infty }f\left ( n \right )$$ equals


ANSWER

$$4/e$$


SOLUTION
Let $$\displaystyle \lim _{ n\rightarrow \infty  }{ f\left( n \right)  } =A=\lim _{ n\rightarrow \infty  }{ { \left[ \left( 1+\frac { 1 }{ n }  \right) \left( 1+\frac { 2 }{ n }  \right) ...\left( 1+\frac { n }{ n }  \right)  \right]  }^{ \dfrac { 1 }{ n }  } } $$

$$\displaystyle \Rightarrow \log { A= } \lim _{ n\rightarrow \infty  }{ \frac { 1 }{ n }  } \sum _{ r=1 }^{ n }{ \log { \left( 1+\frac { r }{ n }  \right)  }  } =\int _{ 0 }^{ 1 }{ \log { \left( 1+x \right)  }  } dx=\log { \frac { 4 }{ e }  } $$

$$\displaystyle \Rightarrow A=\frac { 4 }{ e } $$
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Single Correct Medium Published on 17th 09, 2020
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