Question

$\underset{n\to \infty }{lim}n[\frac{1}{(n+1)(n+2)}+\frac{1}{(n+2)(n+4)}+\cdots +\frac{1}{6n^2}]$ is equal to

• A. $\log \left(\frac{3}{2}\right)$
• B. $\log \left(\frac{5}{2}\right)$
• C. $\log \left(\frac{1}{2}\right)$
• D. $\log \left(\frac{7}{4}\right)$

Answer 1

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

Let $L=\underset{n\to \infty }{lim}n[\frac{1}{(n+1)(n+2)}+\frac{1}{(n+2)(n+4)}+\cdots +\frac{1}{6n^2}]$

$=\underset{n\to \infty }{lim}n\cdot \sum _{r=1}^n\frac{1}{(n+r)(n+2r)}$

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

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

$={\int }_0^1\frac{dx}{(1+x)(1+2x)}={\int }_0^1(\frac{-1}{1+x}+\frac{2}{1+2x})dx$

$={[-\log (1+x)+\log (1+2x)]}_0^1$

$=[-\log 2+\log 3-(-\log 1+\log 1)]=\log \left(\frac{3}{2}\right)$