lim _{n rightarrow infty} frac{(2n(2n-1) dots | vedclass

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(D) Let $L = \lim _{n \rightarrow \infty} \frac{((2n)!/n!)^{1/n}}{n} = \lim _{n \rightarrow \infty} \left( \frac{(2n)!}{n! n^n} \right)^{1/n}$.Taking

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$\int_0^1 \ln (1+x) \, dx$

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(D) Let $L = \lim _{n \rightarrow \infty} \frac{((2n)!/n!)^{1/n}}{n} = \lim _{n \rightarrow \infty} \left( \frac{(2n)!}{n! n^n} \right)^{1/n}$.Taking the natural logarithm on both sides:$\ln L = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \ln \left( \frac{n+k}{n} \right) = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \ln \left( 1 + \frac{k}{n} \right)$.This is a Riemann sum for the integral $\int_0^1 \ln(1+x) \, dx$.Thus,$\ln L = \int_0^1 \ln(1+x) \, dx$,which implies $L = e^{\int_0^1 \ln(1+x) \, dx}$.The expression given in the limit is equivalent to $\int_0^1 \ln(1+x) \, dx$ when evaluated using the Riemann sum definition of the definite integral.

$\lim _{n \rightarrow \infty} \frac{(2n(2n-1) \dots (n+1))^{1/n}}{n} = $

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