Mathematics

# $n\overset{Lt}{\rightarrow}\infty \displaystyle \{\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{3n}\}=$

log 3

##### SOLUTION

$n\overset{Lt}{\rightarrow}\infty \left [ \dfrac{1}{n}+\dfrac{1}{n+1}+............ \dfrac{1}{3n} \right ]$
$n\overset{Lt}{\rightarrow}\infty\sum_{r=0}^{2n}\dfrac{1}{n+r}=n\overset{Lt}{\rightarrow}\infty\sum_{r=0}^{2n}\dfrac{1}{n}\left ( \dfrac{1}{1+\dfrac{r}{n}}\right )$
It is in the form of
$n\overset{Lt}{\rightarrow}\infty\sum_{r=0}^{n}\dfrac{1}{n}f(\dfrac{r}{n})=\int_{0}^{1}f(x)dx$
$n\overset{Lt}{\rightarrow}\infty\sum_{r=0}^{2n}\dfrac{1}{n}\left ( \dfrac{1}{1+\dfrac{r}{n}} \right )=\int_{0}^{2}\dfrac{1}{1+x}dx$
$=log (1+x)\int_{0}^{2}$
$=log 3$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

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