Question

How do you determine if ${{a}_{n}}=\left( 1-\dfrac{1}{8} \right)+\left( \dfrac{1}{8}-\dfrac{1}{27} \right)+\left( \dfrac{1}{27}-\dfrac{1}{64} \right)+...+\left( \dfrac{1}{{{n}^{3}}}-\dfrac{1}{{{\left( n+1 \right)}^{3}}} \right)+...$ converge and find the sum when they exist?

Answer 1

Hint: In this question, we have to find whether the given series is convergent or not. Thus, we will apply the sum of partial numbers to get the solution. First, we will open the brackets of the given series, and then apply the mathematical property which states that the same terms with opposite signs cancel out each other. After that, we will apply the limit on both sides and then apply the limit formula $\underset{n\to \infty }{\mathop{\lim }}\,\left( a-b \right)=\underset{n\to \infty }{\mathop{\lim }}\,\left( a \right)-\underset{n\to \infty }{\mathop{\lim }}\,\left( b \right)$ to get the required solution for the problem.

Complete step by step answer:
According to the problem, we have to find whether the given series is convergent or not
Thus, we will apply the mathematical rule and the sum of partial terms to get the solution.
The series given to us is ${{a}_{n}}=\left( 1-\dfrac{1}{8} \right)+\left( \dfrac{1}{8}-\dfrac{1}{27} \right)+\left( \dfrac{1}{27}-\dfrac{1}{64} \right)+...+\left( \dfrac{1}{{{n}^{3}}}-\dfrac{1}{{{\left( n+1 \right)}^{3}}} \right)+...$ -- (1)
Now, we know that the sum of partial terms is denoted by ${{S}_{n}}$ , we get
${{S}_{n}}=\left( 1-\dfrac{1}{8} \right)+\left( \dfrac{1}{8}-\dfrac{1}{27} \right)+\left( \dfrac{1}{27}-\dfrac{1}{64} \right)+...+\left( \dfrac{1}{{{n}^{3}}}-\dfrac{1}{{{\left( n+1 \right)}^{3}}} \right)+...$
First, we will open the brackets of the equation (1), we get
$\Rightarrow {{S}_{n}}=1-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{64}+...+\dfrac{1}{{{n}^{3}}}-\dfrac{1}{{{\left( n+1 \right)}^{3}}}+...$
As we know, the same terms with opposite signs cancel out each other, we get
$\Rightarrow {{S}_{n}}=1-\dfrac{1}{{{\left( n+1 \right)}^{3}}}$
Thus, now we will apply the limits on both sides in the above equation where n tends to infinity, we get
$\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\left( 1-\dfrac{1}{{{\left( n+1 \right)}^{3}}} \right)$
Now, we will apply the limit formula $\underset{n\to \infty }{\mathop{\lim }}\,\left( a-b \right)=\underset{n\to \infty }{\mathop{\lim }}\,\left( a \right)-\underset{n\to \infty }{\mathop{\lim }}\,\left( b \right)$ on the right-hand side in the above equation, we get
$\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,1-\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{{{\left( n+1 \right)}^{3}}}$
Now, we know that the limit of constant is the constant, and putting the value of n as infinity in the denominator, we get
$\Rightarrow S=1-\dfrac{1}{{{\left( \infty +1 \right)}^{3}}}$
On further solving, we get
$\Rightarrow S=1-\dfrac{1}{\infty }$
Also, we know that $\dfrac{1}{\infty }$ is 0, we get
$\Rightarrow S=1-0$
Therefore, we get
$\Rightarrow S=1$
Thus, for the series ${{a}_{n}}=\left( 1-\dfrac{1}{8} \right)+\left( \dfrac{1}{8}-\dfrac{1}{27} \right)+\left( \dfrac{1}{27}-\dfrac{1}{64} \right)+...+\left( \dfrac{1}{{{n}^{3}}}-\dfrac{1}{{{\left( n+1 \right)}^{3}}} \right)+...$ , it is convergent and its sum is equal to 1.

Note:
While solving this problem, do mention all the steps properly to avoid confusion and mathematical error. Remember if the series is not convergent it implies that it does not have a finite limit, that is the limit will be infinity or minus infinity.