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(B) Let $P = \mathop {\lim }\limits_{n 	o \infty } {\left( {\frac{{n!}}{{{n^n}}}} ight)^{1/n}}$.Taking the natural logarithm on both sides:$\log P

Vedclass · Accepted Answer

$1/e$

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(B) Let $P = \mathop {\lim }\limits_{n 	o \infty } {\left( {\frac{{n!}}{{{n^n}}}} ight)^{1/n}}$.Taking the natural logarithm on both sides:$\log P = \mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \log \left( \frac{1 \cdot 2 \cdot 3 \cdot ... \cdot n}{n^n} ight)$$\log P = \mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum_{r=1}^{n} \log \left( \frac{r}{n} ight)$Using the definition of the definite integral as the limit of a sum,$\mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum_{r=1}^{n} f\left( \frac{r}{n} ight) = \int_0^1 f(x) dx$:$\log P = \int_0^1 \log x \, dx$Integrating by parts,$\int \log x \, dx = x \log x - x$:$\log P = [x \log x - x]_0^1 = (1 \cdot \log 1 - 1) - \mathop {\lim }\limits_{x 	o 0^+} (x \log x - x)$Since $\mathop {\lim }\limits_{x 	o 0^+} x \log x = 0$,we have:$\log P = -1 - 0 = -1$Therefore,$P = e^{-1} = \frac{1}{e}$.

$\mathop {\lim }\limits_{n \to \infty } {\left[ {\frac{{n!}}{{{n^n}}}} \right]^{1/n}}$ equals

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