Mathematics

# $lt_{n\rightarrow \infty} \left[\dfrac{1}{na}+\dfrac{1}{na+1}\dfrac{1}{na+2}......+\dfrac{1}{nb}\right]=$

$\log \left( \dfrac { b }{ a } \right)$

##### SOLUTION
$\underset{n\to \infty}{\lim} \left(\dfrac{1}{na}+\dfrac{1}{na+1} + \dfrac{1}{na+2}+ ... \dfrac{1}{nb}\right)$

$\underset{n\to \infty}{\lim} \left(\dfrac{1}{na} + \dfrac{1}{na+1} + \dfrac{1}{na+2} +....+\dfrac{1}{na+n(b-a)}\right)$

$\underset{n\to \infty}{\lim} \dfrac{1}{n}\left(\dfrac{1}{a} + \dfrac{1}{a+\dfrac{1}{n}} + \dfrac{1}{a+\dfrac{2}{n}} ....... \dfrac{1}{a+\dfrac{(b-a)n}{n}} \right)$

this summation can be converted into definite integration

$\dfrac{1}{n} \to dx$

$\dfrac{r}{n}\to x$

$\displaystyle \int_0^{b-a}\dfrac{dx}{a+x} = ln(a+x)|_0^{b-a}$

$= ln\, b-ln \, a$

$=ln\left(\dfrac{b}{a}\right)$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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