lim _{n} {rightarrow infty}left[left(1+frac{1 | vedclass

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

Question

(B) Let $V = \lim _{n \rightarrow \infty}\left[\prod_{k=1}^{n} \left(1+\frac{k^2}{n^2}\right)\right]^{1 / n}$.Taking the natural logarithm on both sid

Vedclass · Accepted Answer

$e^{\frac{\pi}{2}-2}$

Vedclass · Answer

(B) Let $V = \lim _{n ightarrow \infty}\left[\prod_{k=1}^{n} \left(1+\frac{k^2}{n^2}ight)ight]^{1 / n}$.Taking the natural logarithm on both sides:$\log V = \lim _{n ightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} \log \left(1+\frac{k^2}{n^2}ight)$.This is a Riemann sum of the form $\int_{0}^{1} f(x) dx$ where $f(x) = \log(1+x^2)$:$\log V = \int_{0}^{1} \log(1+x^2) dx$.Using integration by parts,$\int \log(1+x^2) dx = x \log(1+x^2) - \int x \cdot \frac{2x}{1+x^2} dx = x \log(1+x^2) - 2 \int \frac{x^2+1-1}{1+x^2} dx = x \log(1+x^2) - 2x + 2 	an^{-1}(x)$.Evaluating from $0$ to $1$:$\log V = [1 \cdot \log(2) - 2(1) + 2 	an^{-1}(1)] - [0] = \log 2 - 2 + 2(\frac{\pi}{4}) = \log 2 - 2 + \frac{\pi}{2}$.Thus,$V = e^{\log 2 - 2 + \frac{\pi}{2}} = 2 e^{\frac{\pi}{2}-2}$.

$\lim _{n}$ ${\rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{4}{n^2}\right)\left(1+\frac{9}{n^2}\right) \ldots \left(1+\frac{n^2}{n^2}\right)\right]^{1 / n}=$

Similar Questions

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

Let $\lim _{n \rightarrow \infty} \sum_{r=1}^{n} \left( \frac{n}{\sqrt{n^4+r^4}} - \frac{2 n r^2}{(n^2+r^2) \sqrt{n^4+r^4}} \right) = \frac{\pi}{k}.$ Using only the principal values of the inverse trigonometric functions,then $k^2$ is equal to:

The value of $\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n (k^2 x)$ is

For $a \in \mathbb{R}$ (the set of all real numbers),$a \neq -1$,if $\lim_{n \to \infty} \frac{1^a + 2^a + \dots + n^a}{(n+1)^{a-1}[(na+1) + (na+2) + \dots + (na+n)]} = \frac{1}{60}$,then $a$ is equal to:

$\lim _{n \rightarrow \infty} \frac{1}{n^2} \sum_{k=1}^{2n} k e^{k/n} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module