Question

Sum of $\dfrac{1}{1.3}+\dfrac{1}{2.5}+\dfrac{1}{3.7}+\dfrac{1}{4.9}.....$ is
1. $2{{\log }_{e}}2-2$
2. $2-{{\log }_{e}}2$
3. $2{{\log }_{e}}4$
4. ${{\log }_{e}}4$

Answer 1

Hint : To solve this you must know the logics of how to solve to find the sum of an infinite series and how to find it using a ${{T}_{n}}$ be the nth number in the series. Then use the logic of partial differentiation to be able to simplify the expression.

Complete step-by-step answer :
Let ${{T}_{n}}$ be the nth term of the series. Therefore we get;
${{T}_{n}}=\dfrac{1}{n(2n+1)}$
Now dividing this using the method of partial differentiation is we get;
$\dfrac{1}{n(2n+1)}=\dfrac{A}{n}+\dfrac{B}{2n+1}$
$A=\underset{n\to 0}{\mathop{\lim }}\,n\left[ \dfrac{1}{n(2n+1)} \right]$
n is both on the numerator and denominator therefore dividing we can simplify the equation as
$A=\underset{n\to 0}{\mathop{\lim }}\,\left[ \dfrac{1}{2n+1} \right]$
Hence , we put the value of n in the limit to find the limit of this expression which we then get on substituting that
$A= \dfrac{1}{0+1}$
Therefore the value of A will be $1$. Now using this same method of limits to find the value of B we get
$B=\underset{2n+1\to 0}{\mathop{\lim }}\,(2n+1)\left[ \dfrac{1}{2n+1(n)} \right]$
We can write $2n+1\to 0$ as $n\to -\dfrac{1}{2}$ and also the dividing the values of the right side of the equation we get
$B=\underset{2n+1\to 0}{\mathop{\lim }}\,\dfrac{(2n+1)}{(2n+1)}\left[ \dfrac{1}{(n)} \right]$
Cancel $2n+1$ from both the numerator and denominator and we get
$B=\underset{n\to -\dfrac{1}{2}}{\mathop{\lim }}\,\left[ \dfrac{1}{n} \right]$
Now on simplifying the limit by putting the value of n we get
$B=\dfrac{1}{\dfrac{-1}{2}}$
Now multiplying and dividing $-2$ from both numerator and denominator from
$B=\dfrac{1\times -2}{\dfrac{-1}{2}\times -2}$
This gives us
$B=-2$
Now on substituting the value on the main given equation we get;
$\dfrac{1}{n(2n+1)}=\dfrac{1}{n}-\dfrac{2}{2n+1}$
Hence we can use this to find the sum of an infinite series since we know
$S=\sum\limits_{n=1}^{\infty }{{{T}_{n}}}$
Substituting the value of the nth term that we found before
$S=\sum\limits_{n=1}^{\infty }{\dfrac{1}{n}-\dfrac{2}{2n+1}}$
This gives us
$\left( 1-\dfrac{2}{3} \right)+\left( \dfrac{1}{2}-\dfrac{2}{5} \right)+\left( \dfrac{1}{3}-\dfrac{2}{7} \right)+....$
Now we can divide both terms and adding the first term of each bracket and the second term of each bracket we get
$\left( 1+\dfrac{1}{2}+\dfrac{1}{3}+..... \right)-2\left( \dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+... \right)$
Opening the brackets and subtracting we get
$1+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}+....$
This we can also write as
$2-\left( 1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-.... \right)$
Which gives us the sum of the expression that is
$2-{{\log }_{e}}2$
Hence the answer of this question is option 2.
So, the correct answer is “Option 2”.

Note : An infinite series is a series which continues forever until forever and it's realistically impossible to find the end of it. A common mistake made is in the part to find partial differentiation since it's commonly seen that students get confused there.