mathop {lim }limits_{n to infty } {left( {fra | vedclass

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(D) Let $L = \mathop {\lim }\limits_{n 	o \infty } {\left( {\frac{{\left( {n + 1} ight)\left( {n + 2} ight) \ldots \left( {3n} ight)}}{{{n^{2n}

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$\frac{{27}}{{{e^2}}}$

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(D) Let $L = \mathop {\lim }\limits_{n 	o \infty } {\left( {\frac{{\left( {n + 1} ight)\left( {n + 2} ight) \ldots \left( {3n} ight)}}{{{n^{2n}}}}} ight)^{\frac{1}{n}}}$.Taking the natural logarithm on both sides:$\ln L = \mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum\limits_{r=1}^{2n} \ln \left( {\frac{{n+r}}{n}} ight) = \mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum\limits_{r=1}^{2n} \ln \left( {1 + \frac{r}{n}} ight)$.This is a Riemann sum,which converts to the definite integral:$\ln L = \int\limits_0^2 \ln(1+x) dx$.Using integration by parts,$\int \ln(1+x) dx = (1+x)\ln(1+x) - (1+x) + C$.Evaluating the definite integral:$\ln L = \left[ (1+x)\ln(1+x) - x ight]_0^2 = (3\ln 3 - 2) - (1\ln 1 - 0) = 3\ln 3 - 2$.Thus,$L = e^{3\ln 3 - 2} = e^{\ln(3^3) - 2} = e^{\ln 27} \cdot e^{-2} = \frac{27}{e^2}$.

$\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{\left( {n + 1} \right)\left( {n + 2} \right) \ldots \left( {3n} \right)}}{{{n^{2n}}}}} \right)^{\frac{1}{n}}} = $

Similar Questions

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{n} + \frac{1}{{\sqrt {{n^2} + n} }} + \frac{1}{{\sqrt {{n^2} + 2n} }} + \dots + \frac{1}{{\sqrt {{n^2} + (n - 1)n} }}} \right]$ is equal to

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