Question

$n\stackrel{Lt}{\to }\infty \{\frac{1}{n+1}+\frac{1}{n+2}+\dots +\frac{1}{6n}\}=$

• A. log 2
• B. log 3
• C. log 5
• D. log 6

Answer 1

The correct option is D log 6

$n\stackrel{Lt}{\to }\infty {\sum }_{r=1}^{5n}\frac{1}{n+r}=n\stackrel{Lt}{\to }\infty {\sum }_{r=1}^{5n}\frac{1}{n}\left[\frac{1}{1+\left(\frac{r}{n}\right)}\right]$
It is in the form
$n\stackrel{Lt}{\to }\infty {\sum }_{r=1}^n\frac{1}{n}f\left(\frac{r}{n}\right)={\int }_0^{1}f\left(x\right)dx$
So, $n\stackrel{Lt}{\to }\infty {\sum }_{r=1}^{5n}\frac{1}{n}f\left(\frac{r}{n}\right)={\int }_0^5\frac{1}{1+x}dx$
$=log(1+x){\int }_0^5$
$=log\left(6\right)-log\left(1\right)$
$=log6$