Question

Find the value of
$\dfrac{1}{1\times 2}+\dfrac{1}{2\times 3}+\dfrac{1}{3\times 4}+.............$

Answer 1

We solve this question by considering the first term and write the numerator as the difference of the denominators. Then we separate the fractions and write the term as the difference of two fractions. Similarly, we consider and write the second term and third term as differences between the two fractions. Similarly, we can write all the terms as the difference of two fractions and add them. Then we observe that every term will be canceled except the value 1 and we get the required answer.

Complete step-by-step solution
We are asked to find the value of $\dfrac{1}{1\times 2}+\dfrac{1}{2\times 3}+\dfrac{1}{3\times 4}+.............$.
As we see the first term, we can write the numerator “1” as $2-1$.
So, we can write the first term as,
$\begin{aligned} & \Rightarrow \dfrac{1}{1\times 2}=\dfrac{2-1}{1\times 2} \\\ & \Rightarrow \dfrac{1}{1\times 2}=\dfrac{2}{1\times 2}-\dfrac{1}{1\times 2} \\\ & \Rightarrow \dfrac{1}{1\times 2}=\dfrac{1}{1}-\dfrac{1}{2} \\\ & \Rightarrow \dfrac{1}{1\times 2}=1-\dfrac{1}{2}..............\left( 1 \right) \\\ \end{aligned}$
Now let us consider the second term, $\dfrac{1}{2\times 3}$.
We can write the numerator “1” as $3-2$.
So, we can write the second term as,
$\begin{aligned} & \Rightarrow \dfrac{1}{2\times 3}=\dfrac{3-2}{2\times 3} \\\ & \Rightarrow \dfrac{1}{2\times 3}=\dfrac{3}{2\times 3}-\dfrac{2}{2\times 3} \\\ & \Rightarrow \dfrac{1}{2\times 3}=\dfrac{1}{2}-\dfrac{1}{3}............\left( 2 \right) \\\ \end{aligned}$
Now let us consider the third term, $\dfrac{1}{3\times 4}$.
We can write the numerator “1” as $4-3$.
So, we can write the third term as,
$\begin{aligned} & \Rightarrow \dfrac{1}{3\times 4}=\dfrac{4-3}{3\times 4} \\\ & \Rightarrow \dfrac{1}{3\times 4}=\dfrac{4}{3\times 4}-\dfrac{3}{3\times 4} \\\ & \Rightarrow \dfrac{1}{3\times 4}=\dfrac{1}{3}-\dfrac{1}{4}...............\left( 3 \right) \\\ \end{aligned}$
Similarly, we can write every term in the given sum as the difference of numbers in the denominator.
Using the equations (1), (2) and (3) we can write the given sum as,
$\begin{aligned} & \Rightarrow \dfrac{1}{1\times 2}+\dfrac{1}{2\times 3}+\dfrac{1}{3\times 4}+.............=\left( 1-\dfrac{1}{2} \right)+\left( \dfrac{1}{2}-\dfrac{1}{3} \right)+\left( \dfrac{1}{3}-\dfrac{1}{4} \right)+......... \\\ & \Rightarrow \dfrac{1}{1\times 2}+\dfrac{1}{2\times 3}+\dfrac{1}{3\times 4}+.............=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......... \\\ \end{aligned}$
We can see that every number in the above series cancels except 1. So, we get the sum as,
$\begin{aligned} & \Rightarrow \dfrac{1}{1\times 2}+\dfrac{1}{2\times 3}+\dfrac{1}{3\times 4}+.............=1+0+0+0+......... \\\ & \Rightarrow \dfrac{1}{1\times 2}+\dfrac{1}{2\times 3}+\dfrac{1}{3\times 4}+.............=1 \\\ \end{aligned}$
So, we get the value of $\dfrac{1}{1\times 2}+\dfrac{1}{2\times 3}+\dfrac{1}{3\times 4}+.............$ as 1.
Hence the answer is 1.

Note: We can also solve this question in an alternative method by using the sigma notation.
So, we can write the sum $\dfrac{1}{1\times 2}+\dfrac{1}{2\times 3}+\dfrac{1}{3\times 4}+.............$ as,
$\Rightarrow \dfrac{1}{1\times 2}+\dfrac{1}{2\times 3}+\dfrac{1}{3\times 4}+.............=\sum\limits_{n=1}^{\infty }{\dfrac{1}{n\left( n+1 \right)}}$
Then we can write $\dfrac{1}{n\left( n+1 \right)}$ as,
$\begin{aligned} & \Rightarrow \sum\limits_{n=1}^{\infty }{\dfrac{1}{n\left( n+1 \right)}}=\sum\limits_{n=1}^{\infty }{\left( \dfrac{1}{n}-\dfrac{1}{n+1} \right)} \\\ & \Rightarrow \sum\limits_{n=1}^{\infty }{\dfrac{1}{n\left( n+1 \right)}}=\sum\limits_{n=1}^{\infty }{\dfrac{1}{n}}-\sum\limits_{n=1}^{\infty }{\dfrac{1}{n+1}} \\\ \end{aligned}$
Now let us consider the value $\sum\limits_{n=1}^{\infty }{\dfrac{1}{n+1}}$ from the above equation.
Let us assume that $t=n+1$. Then we can write $\sum\limits_{n=1}^{\infty }{\dfrac{1}{n+1}}$ as,
$\begin{aligned} & \Rightarrow \sum\limits_{n=1}^{\infty }{\dfrac{1}{n+1}}=\sum\limits_{t-1=1}^{\infty }{\dfrac{1}{t}} \\\ & \Rightarrow \sum\limits_{n=1}^{\infty }{\dfrac{1}{n+1}}=\sum\limits_{t=1+1}^{\infty }{\dfrac{1}{t}} \\\ & \Rightarrow \sum\limits_{n=1}^{\infty }{\dfrac{1}{n+1}}=\sum\limits_{t=2}^{\infty }{\dfrac{1}{t}} \\\ \end{aligned}$
So, we get that $\sum\limits_{n=1}^{\infty }{\dfrac{1}{n+1}}$ is equal to $\sum\limits_{t=2}^{\infty }{\dfrac{1}{t}}$. Now let us replace $t$ by $n$. Then $\sum\limits_{t=2}^{\infty }{\dfrac{1}{t}}$ becomes $\sum\limits_{n=2}^{\infty }{\dfrac{1}{n}}$.
So, by replacing $\sum\limits_{n=1}^{\infty }{\dfrac{1}{n+1}}$ with $\sum\limits_{n=2}^{\infty }{\dfrac{1}{n}}$ we can write the above equation as,
$\begin{aligned} & \Rightarrow \sum\limits_{n=1}^{\infty }{\dfrac{1}{n\left( n+1 \right)}}=\sum\limits_{n=1}^{\infty }{\dfrac{1}{n}}-\sum\limits_{n=2}^{\infty }{\dfrac{1}{n}} \\\ & \Rightarrow \sum\limits_{n=1}^{\infty }{\dfrac{1}{n\left( n+1 \right)}}=\left( \dfrac{1}{1}+\sum\limits_{n=2}^{\infty }{\dfrac{1}{n}} \right)-\sum\limits_{n=2}^{\infty }{\dfrac{1}{n}} \\\ & \Rightarrow \sum\limits_{n=1}^{\infty }{\dfrac{1}{n\left( n+1 \right)}}=1+\sum\limits_{n=2}^{\infty }{\dfrac{1}{n}}-\sum\limits_{n=2}^{\infty }{\dfrac{1}{n}} \\\ & \Rightarrow \sum\limits_{n=1}^{\infty }{\dfrac{1}{n\left( n+1 \right)}}=1 \\\ \end{aligned}$
So, we get the value of given sum as,
$\Rightarrow \dfrac{1}{1\times 2}+\dfrac{1}{2\times 3}+\dfrac{1}{3\times 4}+.............=\sum\limits_{n=1}^{\infty }{\dfrac{1}{n\left( n+1 \right)}}=1$
Hence the answer is 1.