mathop {lim }limits_{n to infty } left( {frac | vedclass

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Question

(C) The given expression can be written as: $\mathop {\lim }\limits_{n 	o \infty } \sum\limits_{r = 1}^{n} {\frac{{n + r}}{{{n^2} + {r^2}}}} $Divide

Vedclass · Accepted Answer

$\frac{\pi }{4} + \frac{1}{2}\ln 2$

Vedclass · Answer

(C) The given expression can be written as: $\mathop {\lim }\limits_{n 	o \infty } \sum\limits_{r = 1}^{n} {\frac{{n + r}}{{{n^2} + {r^2}}}} $Divide the numerator and denominator by $n^2$: $= \mathop {\lim }\limits_{n 	o \infty } \sum\limits_{r = 1}^{n} {\frac{{\frac{1}{n} + \frac{r}{n^2}}}{{1 + (\frac{r}{n})^2}}} = \mathop {\lim }\limits_{n 	o \infty } \sum\limits_{r = 1}^{n} {\frac{{1 + \frac{r}{n}}}{{1 + (\frac{r}{n})^2}} \cdot \frac{1}{n}}$Using the definition of the definite integral as the limit of a sum,where $x = \frac{r}{n}$ and $dx = \frac{1}{n}$:$= \int_{0}^{1} \frac{1 + x}{1 + x^2} dx$Split the integral:$= \int_{0}^{1} \frac{1}{1 + x^2} dx + \int_{0}^{1} \frac{x}{1 + x^2} dx$Evaluate the integrals:$= [	an^{-1}(x)]_{0}^{1} + \frac{1}{2} [\ln(1 + x^2)]_{0}^{1}$$= (	an^{-1}(1) - 	an^{-1}(0)) + \frac{1}{2} (\ln(2) - \ln(1))$$= \frac{\pi}{4} + \frac{1}{2} \ln 2$

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

Similar Questions

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

$\lim _{n \rightarrow \infty}\left(\frac{n^{2}}{\left(n^{2}+1\right)(n+1)}+\frac{n^{2}}{\left(n^{2}+4\right)(n+2)}+\frac{n^{2}}{\left(n^{2}+9\right)(n+3)}+\ldots+\frac{n^{2}}{\left(n^{2}+n^{2}\right)(n+n)}\right)$ is equal to

By the definition of the definite integral,the value of $\lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{n^2-1^2}}+\frac{1}{\sqrt{n^2-2^2}}+\ldots+\frac{1}{\sqrt{n^2-(n-1)^2}}\right)$ is equal to

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

$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {\frac{1}{n}{e^{\frac{r}{n}}}} $ is

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