Find ${\lim }_{n\to \infty }\left(\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\cdots +\frac{1}{3n}\right)$

The series can be written in sigma notation as:
$$I=\underset{n\to \infty }{\lim }\sum _{r=0}^{2n}\frac{1}{n+r}=\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{r=0}^{2n}\frac{1}{1+\frac{r}{n}}$$
Converting the limit of sum into a definite integral where $\frac{r}{n}\to x$ and $\frac{1}{n}\to dx$:Lower limit: ${\lim }_{n\to \infty }\frac{0}{n}=0$ Upper limit: ${\lim }_{n\to \infty }\frac{2n}{n}=2$ $
$I={\int }_0^2\frac{1}{1+x}dx=[\log (1+x){]}_0^2$ $
$$I=\log (3)-\log (1)=\log 3$$