Question

Let $S_n=\frac{n}{(n+1)(n+2)}+\frac{n}{(n+2)(n+4)}+\frac{n}{(n+3)(n+6)}+\dots \dots \dots +\frac{1}{6n}$ then $\underset{n\to \infty }{\mathrm{Lim}}{S}_n$ has the value equal to

• A. $\ln \frac{3}{2}$
• B. $\ln \frac{9}{2}$
• C. $2\ln \frac{3}{2}$
• D. $\frac{1}{2}\ln \frac{3}{2}$

Answer 1

Answer: (A)

$T_r=\frac{n}{(n+r)(n+2r)}=\frac{1}{n\left(1+\frac{r}{n}\right)\left(1+\frac{2r}{n}\right)}$
$S=\underset{n\to \infty }{\mathrm{Lim}}\sum _{r=1}^n\left(1+\frac{r}{n}\right)\left(1+\frac{2r}{n}\right)=\underset{0}{\overset{1}{\int }}\frac{dx}{(1+x)(1+2x)}=\underset{0}{\overset{1}{\int }}\frac{2dx}{(2+2x)(1+2x)}$
$=2\underset{0}{\overset{1}{\int }}\frac{(2+2x)-(1+2x)}{(2+2x)(1+2x)}dx=2\left[\underset{0}{\overset{1}{\int }}\left(\frac{1}{1+2x}-\frac{1}{2(1+x)}\right)dx\right]={2\cdot \frac{1}{2}\ln (1+2x)|}_0^1-{\ln (1+x)|}_0^1$
$=\ln 3-\ln 2=\ln \frac{3}{2}$