Question

The value of ${lim}_{n\to \infty }[\frac{1}{na}+\frac{1}{na+1}+\frac{1}{na+2}+...+\frac{1}{nb}]$ is

• A. $log\left(ab\right)$
• B. $log\left(\frac{a}{b}\right)$
• C. $log\left(\frac{b}{a}\right)$
• D. $-log\left(ab\right)$

Answer 1

The correct option is C $log\left(\frac{b}{a}\right)$

$\underset{n\to \infty }{lim}(\frac{1}{na}+\frac{1}{na+1}+\frac{1}{na+2}-----\frac{1}{nb})$

$\underset{n\to \infty }{lim}(\frac{1}{na}+\frac{1}{na+1}+\frac{1}{na+2}-----\frac{1}{na+(nb-na)})$

$\sum _{r=o}^{(nb-na)}\frac{1}{na+r}$

it can be converted into definite integration

$\sum _{r=o}^{n(b-a)}\frac{1}{n(a+\frac{r}{n})}$

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

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

$\sum \to \int$

${\int }_0^{b-a}\frac{1}{x+a}dx$

$ln(x+a){|}_0^{b-a}$

$lnb-lna=ln\left(\frac{b}{a}\right)$