Question

$\underset{n\to \infty }{lim}\sum _{r=1}^{4n}\frac{1}{n+r}$ is

• A. ${\log }_{e}5$
• B. $0$
• C. ${\log }_{e}4$
• D. none of these

Answer 1

The correct option is A ${\log }_{e}5$

$\underset{n\to \infty }{lim}\sum _{r=1}^{4n}\frac{1}{n+r}$

$={lim}_{n\to \infty }{\sum }_{r=1}^{4n}\frac{1}{n}\frac{1}{1+\frac{r}{n}}$
$=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^{4n}\frac{1}{1+\frac{r}{n}}$

$={\int }_0^4\frac{1}{1+x}dx={\left[\log (1+x)\right]}_0^4=\log 5$