Question

The value of $\underset{n\to \infty }{lim}[\frac{n}{(n+1)(n+2)}+\frac{n}{(n+2)(n+4)}+\cdots +\frac{1}{6n}]$ is:

• A. $\ln 2$
• B. $\ln \frac{3}{5}$
• C. $\ln 3$
• D. $\ln \frac{3}{2}$

Answer 1

The correct option is D $\ln \frac{3}{2}$

The given limit is
$L=\underset{n\to \infty }{lim}\sum _{r=1}^{n}n.\frac{1}{(n+r)(n+2r)}=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^n\frac{1}{(1+\frac{r}{n})(1+\frac{2r}{n})}$
$=\underset{0}{\overset{1}{\int }}\frac{dx}{(1+x)(1+2x)}$
$=\underset{0}{\overset{1}{\int }}(\frac{-1}{1+x}+\frac{2}{1+2x})dx$
$={[-\ln |1+x|+\ln |1+2x|]}_0^1$
$=\left[\right(-\ln 2+\ln 3)-(-\ln 1+\ln 1\left)\right]$
$=\ln \frac{3}{2}$