Question

How do you find the formula for the nth partial sum of series $\left( \dfrac{5}{1\times 2} \right)+\left( \dfrac{5}{2\times 3} \right)+\left( \dfrac{5}{3\times 4} \right)+......+\left( \dfrac{5}{n\times \left( n+1 \right)} \right)+...$ and use it to find the series sum if the series converges?

Answer 1

Hint : In the given question, we have been asked to find the partial sum of the infinite series. First we need to write the given series in a stand format and then taking out the common constant part out and separating the term into the sum of two terms. Later solving them and substituting in the given series we will then take the limit solving the sum of the nth series. In this way we will get the required answer.

Complete step-by-step answer :
We have given that,
$\left( \dfrac{5}{1\times 2} \right)+\left( \dfrac{5}{2\times 3} \right)+\left( \dfrac{5}{3\times 4} \right)+......+\left( \dfrac{5}{n\times \left( n+1 \right)} \right)+...$
Rewrite the above series in a simpler form, we will get
$\Rightarrow \sum\limits_{1}^{\infty }{\dfrac{5}{n\left( n+1 \right)}}$
Taking out 5 as a common term among all the numbers, we will get
$\Rightarrow 5\left( \sum\limits_{1}^{\infty }{\dfrac{1}{n\left( n+1 \right)}} \right)$
Separating the term $\dfrac{1}{n\left( n+1 \right)}$ into the sum of two terms;
$\dfrac{1}{n\left( n+1 \right)}=\dfrac{A}{n}+\dfrac{B}{n+1}=\dfrac{A\left( n+1 \right)+Bn}{n\left( n+1 \right)}=\dfrac{n\left( A+B \right)+A}{n\left( n+1 \right)}$
Thus,
$\dfrac{1}{n\left( n+1 \right)}=\dfrac{n\left( A+B \right)+A}{n\left( n+1 \right)}$
Cancelling out the common terms, we will get
$1=n\left( A+B \right)+A$
Now,
As we know that,
For the identity principle of polynomials;
$n\left( A+B \right)+A=1$
Thus,
$\Rightarrow A+B=0$
$\Rightarrow A=1$
Solving the above equation, we will get
$\Rightarrow B=-1$
So,
$\dfrac{1}{n\left( n+1 \right)}=\dfrac{1}{n}-\dfrac{1}{n+1}$
We have
$\Rightarrow 5\left( \sum\limits_{1}^{\infty }{\dfrac{1}{n\left( n+1 \right)}} \right)$
Substituting the values, we will get
$\Rightarrow 5\sum\limits_{1}^{\infty }{\left( \dfrac{1}{n}-\dfrac{1}{n+1} \right)}$
Now,
Taking the sum of all the terms in the series;
We have,
${{S}_{n}}=5\left( \left( \dfrac{1}{1}-\dfrac{1}{2} \right)+\left( \dfrac{1}{2}-\dfrac{1}{3} \right)+\left( \dfrac{1}{3}-\dfrac{1}{4} \right)+.....+\left( \dfrac{1}{n}-\dfrac{1}{n+1} \right) \right)$
Here, we can observe that all the terms got canceled by each other except the first term and the last term.
Thus,
${{S}_{n}}=5\left( 1-\dfrac{1}{n+1} \right)$
Now,
Taking the limit i.e. infinity;
$\underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\left( 5\left( 1-\dfrac{1}{n+1} \right) \right)$
We know that dividing a very small number i.e. 1 by the infinity it will give us 0.00000….
Thus,
$\underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\left( 5\left( 1-\dfrac{1}{n+1} \right) \right)=5\left( 1-0 \right)=5$
Hence, this is the required answer.
So, the correct answer is “5”.

Note : While solving these types of questions, students must need to know about the concept of partial sum of the infinite series. A partial sum ${{S}_{n}}$ of an infinite series $\sum\limits_{i=1}^{n}{{{a}_{i}}}$ is the sum of the first ‘n’ terms that is, ${{S}_{n}}={{a}_{1}}+{{a}_{2}}+{{a}_{3}}+....+{{a}_{n}}=\sum\limits_{i=1}^{n}{{{a}_{i}}}$ . The sequence of partial sums of an infinite series is a sequence created by taking, in order 1) the first term, 2) the sum of first two terms, 3) the sum of the first three terms, etc.……, and goes on.