If lim _{n rightarrow infty} frac{1}{n} log l | vedclass

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

Question

(C) Let $l = \lim _{n \rightarrow \infty} \frac{1}{n} \log \left(\frac{(2 n)!}{n^n n!}\right)$.We can write $\frac{(2n)!}{n! n^n} = \frac{(n+1)(n+2)\d

Vedclass · Accepted Answer

$\log x$

Vedclass · Answer

(C) Let $l = \lim _{n \rightarrow \infty} \frac{1}{n} \log \left(\frac{(2 n)!}{n^n n!}\right)$.We can write $\frac{(2n)!}{n! n^n} = \frac{(n+1)(n+2)\dots(n+n)}{n^n} = \prod_{r=1}^n \left(\frac{n+r}{n}\right) = \prod_{r=1}^n \left(1 + \frac{r}{n}\right)$.Taking the logarithm,we get $\log \left(\prod_{r=1}^n \left(1 + \frac{r}{n}\right)\right) = \sum_{r=1}^n \log \left(1 + \frac{r}{n}\right)$.Thus,$l = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \log \left(1 + \frac{r}{n}\right)$.Using the definition of a definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n g\left(\frac{r}{n}\right) = \int_0^1 g(x) dx$.Here,$g(x) = \log(1+x)$,so $l = \int_0^1 \log(1+x) dx$.Let $1+x = t$,then $dx = dt$. When $x=0, t=1$ and when $x=1, t=2$.So,$l = \int_1^2 \log t dt = \int_1^2 \log x dx$.Comparing this with $\int_1^2 f(x) dx$,we get $f(x) = \log x$.

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 $f(x)=$

Similar Questions

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{n} + \frac{1}{{n + 1}} + \frac{1}{{n + 2}} + \dots + \frac{1}{{2n}}} \right] = $

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

$\mathop {\lim }\limits_{n \to \infty } \frac{{{1^p} + {2^p} + {3^p} + ..... + {n^p}}}{{{n^{p + 1}}}} = $

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

Evaluate the limit: $\mathop {\lim}\limits_{n \to \infty } \frac{\pi }{2n} \left( 1 + \cos \frac{\pi }{2n} + \cos \frac{2\pi }{2n} + \dots + \cos \frac{(n - 1)\pi }{2n} \right)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module