Question

$n\stackrel{Lt}{\to }\infty \{\frac{1}{2n+1}+\frac{1}{2n+2}+\frac{1}{2n+3}\dots +\frac{1}{2n+n}\}$

• A. ${\log }_e\left(\frac{1}{3}\right)$
• B. ${\log }_e\left(\frac{2}{3}\right)$
• C. ${\log }_e\left(\frac{3}{2}\right)$
• D. ${\log }_e\left(\frac{4}{3}\right)$

Answer 1

The correct option is C ${\log }_e\left(\frac{3}{2}\right)$

$n\stackrel{Lt}{\to }\infty {\sum }_{r=1}^n\frac{1}{2n+r}=n\stackrel{Lt}{\to }\infty {\sum }_{r=1}^n\frac{1}{n}\frac{1}{(2+\frac{r}{n})}$

It is in the form of
$n\stackrel{Lt}{\to }\infty {\sum }_{r=1}^n\frac{1}{n}f\left(\frac{r}{n}\right)={\int }_0^{1}f\left(x\right)dx$
${\int }_0^1\frac{1}{(2+x)}dx=log(x+2){\int }_0^1$
$=log\left(3\right)-log\left(2\right)$
$=log\left(\frac{3}{2}\right)$