Questions & Answers
Question
Answers
Related Questions
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 one
d)Less than two
Answer
![Verified]()
Verified
Verified
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.
${{S}_{n}}=\dfrac{n}{\left( n+1 \right)\left( n+2 \right)}+\dfrac{n}{\left( n+2 \right)\left( n+4 \right)}+\dfrac{n}{\left( n+3 \right)\left( n+6 \right)}+\ldots +\dfrac{1}{6n}$
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.
×
Sorry!, This page is not available for now to bookmark.
Like this
Liked
-
LIVECourses
-
Crash Courses 2021
-
Vedantu Pro
-
Vedantu Assist
-
Class 12
-
-
JEE
-
JEE Main & Advanced 2021
-
JEE Main & Advanced 2022
-
JEE Main & Advanced 2023
-
-
NEET
-
NEET 2021
-
NEET 2022
-
NEET 2023
-
-
NDA
-
CBSE
-
Commerce
-
Class 11
-
Class 12
-
-
ICSE
-
Maharashtra Board
-
Micro Courses
-
-
Reading & Spoken English -
FREEMaster Classes
-
PREMIUM1-on-1 Classes
-
Study Material
-
NCERT Solutions
-
Previous Year Papers
-
Mock Tests
-
Sample Papers
-
Reference Book Solutions
-
ICSE Solutions
-
School Syllabus
-
Revision Notes
-
Important Questions
-
Math Formula Sheets
-
-
More
Book your FREE Online counselling session
Share your contact information
Your FREE Online counselling session has been booked!
Vedantu academic counsellor will be calling you shortly for your Online Counselling session.
DONE
Please try again later
OK
NCERT Book Solutions
Reference Book Solutions
Toll Free:
1800-120-456-456
Mobile:
+91 988-660-2456
Toll Free:
1800-120-456-456
Mobile:
+91 988-660-2456
(Mon-Sun: 9am - 11pm IST)
Questions?
Chat with us
