Question

The value of $\lim _{n \rightarrow \infty}\left[\frac{1}{n a}+\frac{1}{n a+1}+\frac{1}{n a+2}+\cdots+\frac{1}{n b}\right]$ is

• A. $\log (a b)$
• B. $\log (a / b)$
• C. $\log (b / a)$
• D. $-\log (a b)$

Answer 1

Answer: (C)

The given limit
$L=\displaystyle\lim _{n \rightarrow \infty}\left[\frac{1}{n a}+\frac{1}{n a+1}+\frac{1}{n a+2}+\cdots+\frac{1}{n a+n(b-a)}\right]$
$=\displaystyle\lim _{n \rightarrow \infty} \displaystyle\sum_{r=0}^{(b-a) n} \frac{1}{n a+r}$
$=\displaystyle\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{(b-a) n} \frac{1}{a+r / n}$
$=\int\limits_{0}^{(b-a)} \frac{d x}{a+x}=[\log (a+x)]_{0}^{b-a}$
$=\log b-\log a=\log (b / a)$