If U_{n}=left(1+frac{1^{2}}{n^{2}}right)^{1}l | vedclass

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

Question

(A) Given $U_{n}=\prod_{r=1}^{n}\left(1+\frac{r^{2}}{n^{2}}\right)^{r}$.Let $L=\lim _{n \rightarrow \infty}\left(U_{n}\right)^{-4 / n^{2}}$.Taking nat

Vedclass · Accepted Answer

$\frac{e^{2}}{16}$

Vedclass · Answer

(A) Given $U_{n}=\prod_{r=1}^{n}\left(1+\frac{r^{2}}{n^{2}}\right)^{r}$.Let $L=\lim _{n \rightarrow \infty}\left(U_{n}\right)^{-4 / n^{2}}$.Taking natural logarithm on both sides:$\log L=\lim _{n \rightarrow \infty} \frac{-4}{n^{2}} \sum_{r=1}^{n} r \log \left(1+\frac{r^{2}}{n^{2}}\right)$.We can rewrite this as:$\log L=\lim _{n \rightarrow \infty} \sum_{r=1}^{n}-4 \left(\frac{r}{n}\right) \log \left(1+\left(\frac{r}{n}\right)^{2}\right) \cdot \frac{1}{n}$.This is a Riemann sum,which converts to the definite integral:$\log L = -4 \int_{0}^{1} x \log (1+x^{2}) \, dx$.Let $t = 1+x^{2}$,then $dt = 2x \, dx$.When $x=0, t=1$; when $x=1, t=2$.$\log L = -2 \int_{1}^{2} \log (t) \, dt = -2 [t \log t - t]_{1}^{2}$.$\log L = -2 [(2 \log 2 - 2) - (1 \log 1 - 1)] = -2 [2 \log 2 - 2 + 1] = -2 [2 \log 2 - 1]$.$\log L = -4 \log 2 + 2 = \log (2^{-4}) + \log (e^{2}) = \log \left(\frac{e^{2}}{16}\right)$.Therefore,$L = \frac{e^{2}}{16}$.

If $U_{n}=\left(1+\frac{1^{2}}{n^{2}}\right)^{1}\left(1+\frac{2^{2}}{n^{2}}\right)^{2} \ldots\left(1+\frac{n^{2}}{n^{2}}\right)^{n}$,then $\lim _{n \rightarrow \infty}\left(U_{n}\right)^{\frac{-4}{n^{2}}}$ is equal to :

Similar Questions

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:

The value of $\lim _{n \rightarrow \infty} \left( \frac{1}{\sqrt{4n^2-1}} + \frac{1}{\sqrt{4n^2-4}} + \dots + \frac{1}{\sqrt{4n^2-n^2}} \right)$ is

If $k \in N$ then $\lim _{n \rightarrow \infty}\left[\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\ldots+\frac{1}{k n}\right]=$

$\mathop {\lim }\limits_{n \to \infty } \frac{{1 + {2^4} + {3^4} + .... + {n^4}}}{{{n^5}}} - \mathop {\lim }\limits_{n \to \infty } \frac{{1 + {2^3} + {3^3} + .... + {n^3}}}{{{n^5}}} = $

Evaluate the following definite integral as the limit of a sum: $\int_{-1}^{1} e^{x} dx$

Vedclass Test Series

Exam Paper Generator

Online Exam Module