Question

The value of $\underset{n\to \infty }{lim}[\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n}]$ is

• A. $0$
• B. ${\log }_{e}4$
• C. ${\log }_{e}3$
• D. ${\log }_{e}2$

Answer 1

The correct option is C ${\log }_{e}3$

Expression is $\underset{n\to \infty }{lim}[\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n}]$

We know that $\underset{n\to \infty }{lim}[\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n}]=\underset{n\to \infty }{lim}\left\{\frac{1}{n}[\frac{1}{1}+\frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}+...+\frac{1}{1+\frac{2n}{n}}]\right\}$ $=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=0}^{2n}\frac{1}{(1+\frac{r}{n})}.$

Substituting $\left(\frac{r}{n}\right)=x$ and $\left(\frac{1}{n}\right)=dx.$

We also know that when $r=0,$ then $x=0$.

And when $r=2n,$ then $x=2.$

Therefore, $I={\int }_0^2\frac{1}{1+x}dx={\left[{\log }_e(1+x)\right]}_0^2={\log }_{e}3-{\log }_{e}1$

$\Rightarrow I={\log }_{e}3.$