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\lim_{n\rightarrow \infty }\left ( \frac{(n+1)(n+2).....3n}{n^{2n}} \right )^{\frac{1}{n}}is equal to
Option: 1 \frac{18}{e^{4}}
Option: 2 \frac{27}{e^{2}}
Option: 3 3\log ^{3-2}
Option: 4 3\log ^{3-2}
 

Answers (2)

best_answer

As we learnt

 

Walli's Method -

 

Definite integral by first principle

\int_{a}^{b}f(x)dx= \left ( b-a \right )\lim_{n \to \infty }\frac{1}{n}\left [ f(a) +f(a+h)+f(a+2h)....\right ]

where

h=\frac{b-a}{n}

- wherein

 

 y=\lim_{n\rightarrow \infty }\left [ \frac{(n+1)(n+2)...(n+2n)}{n^{2n}} \right ]^{1/n}

=\lim_{n\rightarrow \infty }\left [ \left ( 1+\frac{1}{n} \right )\left ( 1+\frac{2}{n} \right ) ...\left ( 1+\frac{2n}{n} \right )\right ]^{1/n}

\Rightarrow \log y=\lim_{n\rightarrow \infty }\frac{1}{n}\left [ \left ( 1+\frac{1}{n} \right )\left ( 1+\frac{2}{n} \right ) ...\left ( 1+\frac{2n}{n} \right )\right ]

=\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{r=1}^{2n}\log\left ( 1+\frac{r}{n} \right )

=\int_{0}^{2}\log\left ( 1+x \right )dx

=\left [ \log(1+x).x \right ]-\int_{0}^{2}\frac{1}{1+x}.xdx

= \left [ x\log(1+x)-x +\log(1+x)\right ]^{2}_{0}

=2\log 3-2+\log 2

\Rightarrow y=\frac{27}{e^{2}}

 

 

Posted by

avinash.dongre

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\lim_{n\rightarrow \infty }\left ( \frac{(n+1)(n+2).....3n}{n^{2n}} \right )^{\frac{1}{n}}is equal to
Option: 1 \frac{18}{e^{4}}
Option: 2 \frac{27}{e^{2}}
Option: 3 3\log ^{3-2}
Option: 4 3\log ^{3-2}
 

Posted by

Divya Sharma

View full answer

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