Question

Let $S_n=\frac{n}{(n+1)(n+2)}+\frac{n}{(n+2)(n+4)}+\frac{n}{(n+3)(n+6)}+\ldots \ldots \ldots+\frac{1}{6 n}$ then $\underset{n \rightarrow \infty}{\text{Lim}} S_n$ 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)

$T_r=\frac{n}{(n+r)(n+2 r)}=\frac{1}{n\left(1+\frac{r}{n}\right)\left(1+\frac{2 r}{n}\right)}$
$S=\underset{n \rightarrow \infty}{\text{Lim}} \displaystyle\sum_{r=1}^n\left(1+\frac{r}{n}\right)\left(1+\frac{2 r}{n}\right)=\int\limits_0^1 \frac{d x}{(1+x)(1+2 x)}=\int\limits_0^1 \frac{2 d x}{(2+2 x)(1+2 x)}$
$=2 \int\limits_0^1 \frac{(2+2 x)-(1+2 x)}{(2+2 x)(1+2 x)} d x=2\left[\int\limits_0^1\left(\frac{1}{1+2 x}-\frac{1}{2(1+x)}\right) d x\right]=\left.2 \cdot \frac{1}{2} \ln (1+2 x)\right|_0 ^1-\left.\ln (1+x)\right|_0 ^1$
$=\ln 3-\ln 2=\ln \frac{3}{2}$