Question

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

• A. $\pi /4$
• B. $\pi /6$
• C. log 2
• D. log 3

Answer 1

The correct option is D log 3

$n\stackrel{Lt}{\to }\infty [\frac{1}{n}+\frac{1}{n+1}+............\frac{1}{3n}]$
$n\stackrel{Lt}{\to }\infty {\sum }_{r=0}^{2n}\frac{1}{n+r}=n\stackrel{Lt}{\to }\infty {\sum }_{r=0}^{2n}\frac{1}{n}\left(\frac{1}{1+\frac{r}{n}}\right)$
It is in the form of
$n\stackrel{Lt}{\to }\infty {\sum }_{r=0}^n\frac{1}{n}f\left(\frac{r}{n}\right)={\int }_0^{1}f\left(x\right)dx$
$n\stackrel{Lt}{\to }\infty {\sum }_{r=0}^{2n}\frac{1}{n}\left(\frac{1}{1+\frac{r}{n}}\right)={\int }_0^2\frac{1}{1+x}dx$
$=log(1+x){\int }_0^2$
$=log3$