Question

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

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

Answer 1

Answer: (C)

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