Question

The value of $\underset{n\to \infty }{lim}[\frac{1}{na}+\frac{1}{na+1}+\frac{1}{na+2}+\cdots +\frac{1}{nb}]$ where $a,b>0$ and $a\ne b$ is:

• A. $\ln \left(\frac{b}{a}\right)$
• B. $\ln \left(ab\right)$
• C. $\ln \left(a^{2}b\right)$
• D. $\ln (a+b)$

Answer 1

The correct option is A $\ln \left(\frac{b}{a}\right)$

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