Question

Evaluate: $\underset{n\to \infty }{lim}(\frac{1}{n}+\frac{1}{n+1}+....+\frac{1}{3n})$

• A. $\log 2$
• B. $\log 3$
• C. $\log 5$
• D. $0$

Answer 1

The correct option is B $\log 3$

${lim}_{n\to \infty }(\frac{1}{n}+\frac{1}{n+1}+....+\frac{1}{3+n})$

$={lim}_{n\to \infty }{\sum }_{r=0}^{2n}\frac{1}{n+r}$

$={lim}_{n\to \infty }{\sum }_{r=0}^{2n}\left(\frac{1}{1+\left(\frac{r}{n}\right)}\right).\frac{1}{n}$

put $\frac{r}{n}=x\Rightarrow \frac{1}{n}=dx$

$={\int }_0^2\frac{1}{1+x}dx$
$={\left[\log (1+x)\right]}_0^2$

$=\log 3-\log 1$
$=\log 3$