Question

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

• A. $\log \left(2/3\right)$
• B. $\log \left(3/2\right)$
• C. $\log \left(4/3\right)$
• D. $\log \left(3/4\right)$

Answer 1

The correct option is C $\log \left(4/3\right)$

$\underset{n\to \infty }{lim}{\sum }_{r=1}^n\frac{1}{(3n+r)}=\underset{n\to \infty }{lim}{\sum }_{r=1}^n\frac{1}{n}\left(\frac{1}{3+\left(\frac{r}{n}\right)}\right)$
It is in the form of
$\underset{n\to \infty }{lim}{\sum }_{r=1}^n\frac{1}{n}f\left(\frac{r}{n}\right)={\int }_0^{1}f\left(x\right)dx$
$={\int }_0^1\frac{1}{3+x}dx$
$=\log (3+x{)}_0^1$
$=\log \left(4\right)-log\left(3\right)$
$=\log \frac{4}{3}$