If f(n) = frac{1}{n} [(n+1)(n+2)(n+3) ldots ( | vedclass

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(A) Let $A = \lim_{n \rightarrow \infty} \frac{1}{n} [(n+1)(n+2) \ldots (2n)]^{\frac{1}{n}}$.We can write this as $A = \lim_{n \rightarrow \infty} \le

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$\frac{4}{e}$

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(A) Let $A = \lim_{n \rightarrow \infty} \frac{1}{n} [(n+1)(n+2) \ldots (2n)]^{\frac{1}{n}}$.We can write this as $A = \lim_{n \rightarrow \infty} \left[ \frac{1}{n^n} (n+1)(n+2) \ldots (2n) \right]^{\frac{1}{n}} = \lim_{n \rightarrow \infty} \left[ \left(1+\frac{1}{n}\right) \left(1+\frac{2}{n}\right) \ldots \left(1+\frac{n}{n}\right) \right]^{\frac{1}{n}}$.Taking the natural logarithm on both sides:$\log A = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \log \left(1+\frac{r}{n}\right)$.Using the definition of a definite integral as a limit of a sum:$\log A = \int_{0}^{1} \log(1+x) dx$.Let $u = 1+x$, then $du = dx$. When $x=0, u=1$; when $x=1, u=2$.$\log A = \int_{1}^{2} \log u du = [u \log u - u]_{1}^{2} = (2 \log 2 - 2) - (1 \log 1 - 1) = 2 \log 2 - 2 + 1 = \log 4 - 1$.Since $1 = \log e$, we have $\log A = \log 4 - \log e = \log \left(\frac{4}{e}\right)$.Therefore, $A = \frac{4}{e}$.

If $f(n) = \frac{1}{n} [(n+1)(n+2)(n+3) \ldots (2n)]^{\frac{1}{n}}$, then $\lim_{n \rightarrow \infty} f(n) =$

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