Question

Find the sum of the following series to \[n\] terms and to infinity: $\dfrac{1}{{1 \cdot 2}} + \dfrac{1}{{2 \cdot 3}} + \dfrac{1}{{3 \cdot 4}} + ......$

Answer 1

Hint: The given series $\dfrac{1}{{1 \cdot 2}} + \dfrac{1}{{2 \cdot 3}} + \dfrac{1}{{3 \cdot 4}} + ......$ follows a pattern, i.e., $\dfrac{1}{{n\left( {n + 1} \right)}}$. Therefore, in this case, the ${n^{th}}$ term is ${t_n} = \dfrac{1}{{n\left( {n + 1} \right)}}$. The sum of \[n\] terms of a series is calculate by the formula, ${S_n} = \sum\limits_{n = 1}^n {\left( {{t_n}} \right)} $, Whereas the sum of infinite ($\infty $) terms can be calculated by ${S_\infty } = \sum\limits_{n = 1}^\infty {\left( {{t_n}} \right)} $.

Complete step-by-step answer:
Given series is: $\dfrac{1}{{1 \cdot 2}} + \dfrac{1}{{2 \cdot 3}} + \dfrac{1}{{3 \cdot 4}} + ......$
The given series follows a pattern, i.e., $\dfrac{1}{{n\left( {n + 1} \right)}}$.
Therefore, the general term (${n^{th}}$term) of the given series is, ${t_n} = \dfrac{1}{{n\left( {n + 1} \right)}}$
For converting it into partial fraction, we can rewrite it as, ${t_n} = \dfrac{{n + 1 - n}}{{n\left( {n + 1} \right)}}$
or ${t_n} = \dfrac{{n + 1}}{{n\left( {n + 1} \right)}} - \dfrac{n}{{n\left( {n + 1} \right)}}$
$ \Rightarrow $${t_n} = \dfrac{1}{n} - \dfrac{1}{{\left( {n + 1} \right)}}$ …… (1)
(A) Sum of \[n\] terms of the given series:
The sum of \[n\] terms of a series is given by ,${S_n} = \sum\limits_{n = 1}^n {\left( {{t_n}} \right)} $
Therefore, sum of \[n\] terms is, ${S_n} = \sum\limits_{n = 1}^n {\left( {\dfrac{1}{n} - \dfrac{1}{{n + 1}}} \right)} $ [Using (1)]
$ \Rightarrow $${S_n} = \left( {\dfrac{1}{1} - \dfrac{1}{2}} \right) + \left( {\dfrac{1}{2} - \dfrac{1}{3}} \right) + \left( {\dfrac{1}{3} - \dfrac{1}{4}} \right) + .............. - \dfrac{1}{n} + \left( {\dfrac{1}{n} - \dfrac{1}{{n + 1}}} \right)$
All the terms were cancel out with each other except $\dfrac{1}{1}$ and $ - \dfrac{1}{{n + 1}}$,
$\therefore {S_n} = \dfrac{1}{1} - \dfrac{1}{{n + 1}}$
Now, solving the above equation by taking LCM,
$ \Rightarrow $${S_n} = \dfrac{{n + 1 - 1}}{{n + 1}}$
$ \Rightarrow $${S_n} = \dfrac{n}{{n + 1}}$
Hence, the sum of \[n\] terms of a given series is $\dfrac{n}{{n + 1}}$.
(B) Sum of infinity terms of the given series:
The sum of infinite terms of a series is given by,${S_\infty } = \sum\limits_{n = 1}^\infty {\left( {{t_n}} \right)} $
Therefore, sum of infinite terms is, ${S_\infty } = \sum\limits_{n = 1}^\infty {\left( {\dfrac{1}{n} - \dfrac{1}{{n + 1}}} \right)} $ [Using (1)]
$ \Rightarrow $${S_\infty } = \left( {\dfrac{1}{1} - \dfrac{1}{2}} \right) + \left( {\dfrac{1}{2} - \dfrac{1}{3}} \right) + \left( {\dfrac{1}{3} - \dfrac{1}{4}} \right) + .............. - \dfrac{1}{\infty } + \left( {\dfrac{1}{\infty } - \dfrac{1}{{\infty + 1}}} \right)$
All the terms were cancel out with each other except $\dfrac{1}{1}$ and $\dfrac{1}{{\infty + 1}}$,
$\therefore {S_\infty } = \dfrac{1}{1} - \dfrac{1}{{\infty + 1}}$
Since, $\dfrac{1}{\infty } = 0$$ \Rightarrow \dfrac{1}{{\infty + 1}} = 0$,
$ \Rightarrow $${S_\infty } = 1 - 0$
$ \Rightarrow $${S_\infty } = 1$

Hence, the sum of infinite terms of given series is $1$.

Note: In mathematics, the symbol $\sum {} $ is used for summation. The value below the sigma operator gives us the starting integer, while the top value gives us the upper bound.
For ex- $\sum\limits_{n = 1}^4 {{n^2} = {1^2} + } {2^2} + {3^2} + {4^2} = 1 + 4 + 9 + 16 = 30$.