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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$\frac{4}{e}$

Vedclass · Answer

(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 $\lim_{n \rightarrow \infty} f(n) =$

Similar Questions

$\lim _{n \rightarrow \infty}\left[\frac{1}{n}+\frac{n}{(n+1)^{2}}+\frac{n}{(n+2)^{2}}+\ldots+\frac{n}{(2 n-1)^{2}}\right]$ is equal to ...... .

If $\lim _{n \rightarrow \infty} \frac{1}{n} \log \left(\frac{(2 n)!}{n^n \cdot n!}\right)=\int_1^2 f(x) d x$,then $f(x)=$

$\lim _{n}$ ${\rightarrow \infty}\left(\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\frac{n}{n^2+3^2}+\ldots+\frac{n}{n^2+(2n)^2}\right)=$

$\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}}} = $

$\lim _{n \rightarrow \infty} \frac{1}{n^{k+1}}\left[2^k+4^k+6^k+\ldots+(2 n)^k\right]=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module