If $a$ and $b$ are positive integers such that $b>a,$ then $\underset{n\to \infty }{\lim }\left[\frac{1}{na}+\frac{1}{na+1}+\frac{1}{na+2}+\dots +\frac{1}{nb}\right]=$

We have,

$\underset{n\to \infty }{\lim }\left[\frac{1}{na}+\frac{1}{na+1}+\frac{1}{na+2}+\dots +\frac{1}{nb}\right]$

$=\underset{n\to \infty }{\lim }\left[\frac{1}{na}+\frac{1}{na+1}+\frac{1}{na+2}+\dots +\frac{1}{na+n\left(b-a\right)}\right]$

$=\underset{n\to \infty }{\lim }\left[\frac{1}{na}+\frac{1}{na+1}+\frac{1}{na+2}+\dots +\frac{1}{na+n\left(b-a\right)}\right]$

$=\underset{n\to \infty }{\lim }\frac{1}{n}\left[\frac{1}{a}+\frac{1}{\left(a+{\displaystyle \frac{1}{n}}\right)}+\frac{1}{\left(a+{\displaystyle \frac{2}{n}}\right)}+\dots +\frac{1}{a+{\displaystyle \frac{n\left(b-a\right)}{n}}}\right]$

$=\underset{n\to \infty }{\lim }\sum _{r=0}^{n\left(b-a\right)}\frac{1}{n}\left(\frac{1}{a+{\displaystyle \frac{r}{n}}}\right)$

Now, replacing $\frac{1}{n}\to dx, \frac{r}{n}\to x$ and converting limits, we get

$={\int }_0^{\left(b-a\right)}\left(\frac{dx}{a+x}\right)$

$={\left[\log \left(a+x\right)\right]}_0^{\left(b-a\right)}$

$=\log \left(\frac{b}{a}\right)$.