lim _{n rightarrow infty}left[frac{n+3}{n^2+1 | vedclass

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

Question

(A) The given expression is $S = \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{n+3r}{n^2+r^2}$.Dividing numerator and denominator by $n^2$, we get $

Vedclass · Accepted Answer

$\frac{\pi}{4}+\frac{3}{2} \ln 2$

Vedclass · Answer

(A) The given expression is $S = \lim _{n ightarrow \infty} \sum_{r=1}^n \frac{n+3r}{n^2+r^2}$.Dividing numerator and denominator by $n^2$, we get $S = \lim _{n ightarrow \infty} \sum_{r=1}^n \frac{\frac{1}{n} + 3\frac{r}{n^2}}{1 + (\frac{r}{n})^2} = \lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{1 + 3(\frac{r}{n})}{1 + (\frac{r}{n})^2}$.Using the definition of definite integral as the limit of a sum, we have $\int_0^1 \frac{1+3x}{1+x^2} dx$.This can be split into $\int_0^1 \frac{1}{1+x^2} dx + 3 \int_0^1 \frac{x}{1+x^2} dx$.Evaluating the integrals, we get $[	an^{-1} x]_0^1 + \frac{3}{2} [\ln(1+x^2)]_0^1$.$= (\frac{\pi}{4} - 0) + \frac{3}{2} (\ln 2 - \ln 1) = \frac{\pi}{4} + \frac{3}{2} \ln 2$.

$\lim _{n \rightarrow \infty}\left[\frac{n+3}{n^2+1^2}+\frac{n+6}{n^2+2^2}+\frac{n+9}{n^2+3^2}+\ldots+\frac{2}{n}\right]=$

Similar Questions

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

$\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{{\left( {n + 1} \right)}^{1/3}}}}{{{n^{4/3}}}} + \frac{{{{\left( {n + 2} \right)}^{1/3}}}}{{{n^{4/3}}}} + \dots + \frac{{{{\left( {2n} \right)}^{1/3}}}}{{{n^{4/3}}}}} \right)$ is equal to

$\int_0^3 (2+x^2) dx = $

Evaluate the following definite integral as a limit of sums:
$\int_{2}^{3} x^{2} d x$

$\lim _{n \rightarrow \infty} \frac{\sqrt{1}+\sqrt{2}+\ldots+\sqrt{n-1}}{n \sqrt{n}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module