Question

The value of $\underset{n\to \infty }{lim}\{\frac{1}{na}+\frac{1}{na+1}+\frac{1}{na+2}+\cdots +\frac{1}{na+n(b-a)}\}$ is?

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

Answer 1

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

The given limit
$L=\underset{n\to \infty }{lim}\{\frac{1}{na}+\frac{1}{na+1}+\frac{1}{na+2}+\cdots +\frac{1}{na+n(b-a)}\}$
$=\underset{n\to \infty }{lim}\sum _{r=0}^{n(b-a)}\frac{1}{na+r}$
$=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=0}^{(b-a)n}\frac{1}{a+r/n}$
$={\int }_0^{(b-a)}\frac{dx}{a+x}={\left[\log (a+x)\right]}_0^{b-a}$ $(\frac{r}{n}=x)$
$=\log b-\log a$
$=\log \left(\frac{b}{a}\right)$