If lim _{n rightarrow infty} sum_{r=1}^n frac | vedclass

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

Question

(C) Given the limit expression: $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{4 r^3}{r^4+n^4}=p$. We can rewrite the sum as: $\lim _{n \rightarrow

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Given the limit expression: $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{4 r^3}{r^4+n^4}=p$. We can rewrite the sum as: $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{4 r^3}{n^4 \left(1+\frac{r^4}{n^4}\right)} = \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{4 (r/n)^3}{1+(r/n)^4} \cdot \frac{1}{n}$. Using the definition of a definite integral as the limit of a sum, let $x = \frac{r}{n}$ and $dx = \frac{1}{n}$. As $n \rightarrow \infty$, the sum becomes the integral from $0$ to $1$: $p = \int_0^1 \frac{4x^3}{1+x^4} dx$. Let $t = 1+x^4$, then $dt = 4x^3 dx$. When $x=0$, $t=1$. When $x=1$, $t=2$. Thus, $p = \int_1^2 \frac{dt}{t} = [\ln t]_1^2 = \ln 2 - \ln 1 = \ln 2$. Therefore, $e^p = e^{\ln 2} = 2$.

If $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{4 r^3}{r^4+n^4}=p$, then $e^p=$

Similar Questions

$\lim _{n \rightarrow \infty} \frac{1}{2^{n}}\left(\frac{1}{\sqrt{1-\frac{1}{2^{n}}}}+\frac{1}{\sqrt{1-\frac{2}{2^{n}}}}+\frac{1}{\sqrt{1-\frac{3}{2^{n}}}}+\ldots+\frac{1}{\sqrt{1-\frac{2^{n}-1}{2^{n}}}}\right)$ is equal to

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

The value of $\lim _{n}$ ${\rightarrow \infty} \left( \frac{\sqrt{n}}{\sqrt{n^{3}}}+\frac{\sqrt{n}}{\sqrt{(n+4)^{3}}}+\frac{\sqrt{n}}{\sqrt{(n+8)^{3}}}+\cdots +\frac{\sqrt{n}}{\sqrt{[n+4(n-1)]^{3}}} \right)$ is

$\mathop {\lim }\limits_{n \to \infty } \left( \frac{1^2}{1^3 + n^3} + \frac{2^2}{2^3 + n^3} + \dots + \frac{n^2}{n^3 + n^3} \right)$ is equal to

$\mathop {\lim }\limits_{n \to \infty } {\left\{ {\left( {1 + \frac{{{1^2}}}{{{n^2}}}} \right)\left( {1 + \frac{{{2^2}}}{{{n^2}}}} \right)\left( {1 + \frac{{{3^2}}}{{{n^2}}}} \right) \dots \left( {1 + \frac{{{{(n - 1)}^2}}}{{{n^2}}}} \right)} \right\}^{1/n}}$ equals to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module