Question

# Find the value of $\underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}$ when ${{S}_{n}}=\dfrac{n}{\left( n+1 \right)\left( n+2 \right)}+\dfrac{n}{\left( n+2 \right)\left( n+4 \right)}+\dfrac{n}{\left( n+3 \right)\left( n+6 \right)}+\ldots +\dfrac{1}{6n}$ a)$ln\dfrac{3}{2}$b)$ln\dfrac{9}{2}$c)Greater than oned)Less than two

Consider the given expression,
${{S}_{n}}=\dfrac{n}{\left( n+1 \right)\left( n+2 \right)}+\dfrac{n}{\left( n+2 \right)\left( n+4 \right)}+\dfrac{n}{\left( n+3 \right)\left( n+6 \right)}+\ldots +\dfrac{1}{6n}$
This can be converted to summation as,
${{S}_{n}}=\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{n}{\left( n+r \right)\left( n+2r \right)}$
Now we will apply limits, we get
$\underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{n}{\left( n+r \right)\left( n+2r \right)}$
Dividing numerator and denominator by n2, we get
$\underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{\dfrac{n}{{{n}^{2}}}}{\dfrac{\left( n+r \right)}{n}\dfrac{\left( n+2r \right)}{n}}$
$\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{\dfrac{1}{n}}{\left( \dfrac{n}{n}+\dfrac{r}{n} \right)\left( \dfrac{n}{n}+\dfrac{2r}{n} \right)}$
$\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{\dfrac{1}{n}}{\left( 1+\dfrac{r}{n} \right)\left( 1+\dfrac{2r}{n} \right)}\ldots \ldots \ldots .\left( i \right)$
Now let,
$\dfrac{r}{n}=x\Rightarrow \dfrac{1}{n}=dx$
Let’s find the limits,
When $r=1\Rightarrow x=0$
When $r=n\Rightarrow x=1$
Considering these values the summation can be written as integral form. We get,
$\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{\dfrac{1}{n}}{\left( 1+\dfrac{r}{n} \right)\left( 1+\dfrac{2r}{n} \right)}=\underset{0}{\overset{1}{\mathop \int }}\,\dfrac{dx}{\left( 1+x \right)\left( 1+2x \right)}\ldots \ldots \left( ii \right)$
Now we will apply partial dfraction to simplify the above equation.
$\dfrac{1}{\left( 1+x \right)\left( 1+2x \right)}=\dfrac{A}{1+x}+\dfrac{B}{1+2x}\ldots ..\left( iii \right)$
$\Rightarrow 1=A\left( 1+2x \right)+B\left( 1+x \right)$
Now put (x=-1), we get
$\Rightarrow 1=A\left( 1+2\left( -1 \right) \right)+B\left( 1-1 \right)$
$\Rightarrow 1=A\left( -1 \right)+0$
$\Rightarrow A=-1$
Now put $x=-\dfrac{1}{2}$, we get
$\Rightarrow 1=A\left( 1+2\left( -\dfrac{1}{2} \right) \right)+B\left( 1-\dfrac{1}{2} \right)$
$\Rightarrow 1=A\left( 1-1 \right)+B\left( \dfrac{2-1}{2} \right)$
$\Rightarrow 1=0+B\left( \dfrac{1}{2} \right)$
$\Rightarrow B=2$
Now substituting the value of ‘A’ and ‘B’ in equation (iii), we get
$\dfrac{1}{\left( 1+x \right)\left( 1+2x \right)}=\dfrac{-1}{1+x}+\dfrac{2}{1+2x}$
Substituting this in equation (ii), we get
$\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{\dfrac{1}{n}}{\left( 1+\dfrac{r}{n} \right)\left( 1+\dfrac{2r}{n} \right)}=\underset{0}{\overset{1}{\mathop \int }}\,\left[ \dfrac{2}{1+2x}-\dfrac{1}{1+x} \right]dx$
Applying linearity, we get
$\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{\dfrac{1}{n}}{\left( 1+\dfrac{r}{n} \right)\left( 1+\dfrac{2r}{n} \right)}=2\underset{0}{\overset{1}{\mathop \int }}\,\dfrac{1}{1+2x}dx-\underset{0}{\overset{1}{\mathop \int }}\,\dfrac{1}{1+x}dx$
But we know, $\mathop{\int }^{}\dfrac{1}{u}=\ln \left( u \right)$, so above equation becomes,
$\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{\dfrac{1}{n}}{\left( 1+\dfrac{r}{n} \right)\left( 1+\dfrac{2r}{n} \right)}=2\left[ \ln \left( 1+2x \right)\underset{0}{\overset{1}{\mathop \int }}\,\left( 1+2x \right)dx \right]-\left[ \ln \left( 1+x \right)\underset{0}{\overset{1}{\mathop \int }}\,\left( 1+x \right)dx \right]$
$\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{\dfrac{1}{n}}{\left( 1+\dfrac{r}{n} \right)\left( 1+\dfrac{2r}{n} \right)}=2\left[ \ln \left( 1+2x \right)\left( \dfrac{1}{2} \right) \right]_{0}^{1}-\left[ \ln \left( 1+x \right)\left( 1 \right) \right]_{0}^{1}$
$\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{\dfrac{1}{n}}{\left( 1+\dfrac{r}{n} \right)\left( 1+\dfrac{2r}{n} \right)}=\left[ \ln \left( 1+2x \right) \right]_{0}^{1}-\left[ \ln \left( 1+x \right) \right]_{0}^{1}$
Applying the upper bound and lower bound, we get
$\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{\dfrac{1}{n}}{\left( 1+\dfrac{r}{n} \right)\left( 1+\dfrac{2r}{n} \right)}=\left[ \ln \left( 1+2\left( 1 \right) \right)-\ln \left( 1+2\left( 0 \right) \right) \right]-\left[ \ln \left( 1+\left( 1 \right) \right)-\ln \left( 1+\left( 0 \right) \right) \right]$
$\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{\dfrac{1}{n}}{\left( 1+\dfrac{r}{n} \right)\left( 1+\dfrac{2r}{n} \right)}=\left[ \ln \left( 3 \right)-\ln \left( 1 \right) \right]-\left[ \ln \left( 2 \right)-\ln \left( 1 \right) \right]$
But we know, $\ln 1=0$, so we get
$\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{\dfrac{1}{n}}{\left( 1+\dfrac{r}{n} \right)\left( 1+\dfrac{2r}{n} \right)}=\left[ \ln \left( 3 \right)-0 \right]-\left[ \ln \left( 2 \right)-0 \right]$
. $\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{\dfrac{1}{n}}{\left( 1+\dfrac{r}{n} \right)\left( 1+\dfrac{2r}{n} \right)}=\ln 3-\ln 2$
We know, $\log a-\log b=\log \left( \dfrac{a}{b} \right)$, so the above equation becomes,
$\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{\dfrac{1}{n}}{\left( 1+\dfrac{r}{n} \right)\left( 1+\dfrac{2r}{n} \right)}=\ln \dfrac{3}{2}$
Substituting this value in equation (i), we get
$\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\ln \dfrac{3}{2}$
As the RHS is free of variable, so we can remove the limit, so we get
$\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\ln \dfrac{3}{2}$
Hence, the correct option for the given question is option (a).

Note - The following equation $\underset{0}{\overset{1}{\mathop \int }}\,\dfrac{dx}{\left( 1+x \right)\left( 1+2x \right)}$ can be solved by partial fraction method as well as substitution method. Among both, the substitution method is the easiest one.