Question

Let ${{S}_{n}}=\dfrac{n}{\left( n+1 \right)\left( n+2 \right)}+\dfrac{n}{\left( n+2 \right)\left( n+4 \right)}+\dfrac{n}{\left( n+3 \right)\left( n+6 \right)}+\ldots +\dfrac{1}{6n}$ , then find the value of $\underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}$ is
(a) $ln\dfrac{3}{2}$
(b) $ln\dfrac{9}{2}$
(c) Greater than one
(d) Less than two

Answer 1

Hint: Convert summation to integral using Riemann integral concept and then apply the limits.

Complete step-by-step solution -
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
\[\underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{\dfrac{n}{{{n}^{2}}}}{\dfrac{\left( n+r \right)}{n}\dfrac{\left( n+2r \right)}{n}}\]
In the denominator we will separate the terms, then we get
\[\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{\dfrac{1}{n}}{\left( \dfrac{n}{n}+\dfrac{r}{n} \right)\left( \dfrac{n}{n}+\dfrac{2r}{n} \right)}\]
Cancelling the like terms, we get
\[\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{\dfrac{1}{n}}{\left( 1+\dfrac{r}{n} \right)\left( 1+\dfrac{2r}{n} \right)}\ldots \ldots \ldots .\left( i \right)\]
Now we know the way to solve the summation limit is to convert the summation into integral.
For this first let us assume
\[\dfrac{r}{n}=x\Rightarrow \dfrac{1}{n}=dx\]
Let’s find the limits of integrals,
When \[r=1\], then $\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{r}{n}=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}=0$ , therefore, $x=0$.
When $r=n$ , then $\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{r}{n}=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{n}{n}=1$, therefore, $x=1$.
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 decomposition fraction 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).
So, the answer is Option (a)

Note: In the following equation,
\[\underset{0}{\overset{1}{\mathop \int }}\,\dfrac{dx}{\left( 1+x \right)\left( 1+2x \right)}\]
Instead of solving by partial decomposition fraction method we can use substitution method too. The answer will be the same.
One more possible mistake is when converting the summation to integral, the student will get confused on how to get the limit of integral.