Question

The value of $\mathop {\lim }\limits_{n \to \infty } \dfrac{{\left( {\sum {{n^2}} } \right)\left( {\sum {{n^3}} } \right)}}{{\left( {\sum {{n^6}} } \right)}}$ is
A.1
B.$\dfrac{3}{2}$
C.$\dfrac{5}{6}$
D.$\dfrac{7}{{12}}$

Answer 1

We will use the sum of squares of n natural numbers given by $\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$, the sum of cubes of n natural numbers given by ${\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2}$ and the sum of n natural numbers raise to the power 6 given by $\dfrac{1}{{42}}n\left( {n + 1} \right)\left( {2n + 1} \right)\left( {3{n^4} + 6{n^3} - 3n + 1} \right)$. We will simplify these sum of the series after putting them in the required equation $\mathop {\lim }\limits_{n \to \infty } \dfrac{{\left( {\sum {{n^2}} } \right)\left( {\sum {{n^3}} } \right)}}{{\left( {\sum {{n^6}} } \right)}}$. Upon further simplification, we will get the value of this function and we will check which of the solutions matches the obtained answer.

Complete step-by-step answer:
We are required to find the value of $\mathop {\lim }\limits_{n \to \infty } \dfrac{{\left( {\sum {{n^2}} } \right)\left( {\sum {{n^3}} } \right)}}{{\left( {\sum {{n^6}} } \right)}}$.
We know that the sum of squares of first n natural numbers is given by: $\sum {{n^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}}$
Similarly, the sum of the cubes of n natural numbers is given by: $\sum {{n^3} = } {\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2}$ and the sum of n natural numbers raise to the power 6 is given by:$\sum {{n^6} = } \dfrac{1}{{42}}n\left( {n + 1} \right)\left( {2n + 1} \right)\left( {3{n^4} + 6{n^3} - 3n + 1} \right)$
Substituting these values in $\mathop {\lim }\limits_{n \to \infty } \dfrac{{\left( {\sum {{n^2}} } \right)\left( {\sum {{n^3}} } \right)}}{{\left( {\sum {{n^6}} } \right)}}$, we get
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{{\left( {\sum {{n^2}} } \right)\left( {\sum {{n^3}} } \right)}}{{\left( {\sum {{n^6}} } \right)}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}{{\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)}^2}}}{{\dfrac{1}{{42}}n\left( {n + 1} \right)\left( {2n + 1} \right)\left( {3{n^4} + 6{n^3} - 3n + 1} \right)}}$
Simplifying this equation, we get
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{{\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}{{\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)}^2}}}{{\dfrac{1}{{42}}n\left( {n + 1} \right)\left( {2n + 1} \right)\left( {3{n^4} + 6{n^3} - 3n + 1} \right)}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{\dfrac{1}{6}\left( {2{n^3} + 3{n^2} + n} \right)\dfrac{1}{4}\left( {{n^4} + 2{n^3} + {n^2}} \right)}}{{\dfrac{1}{{42}}\left( {6{n^7} + 21{n^6} + 21{n^5} - 7{n^3} + n} \right)}}$
Upon simplifying the numerator, we get
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{{\left( {\sum {{n^2}} } \right)\left( {\sum {{n^3}} } \right)}}{{\left( {\sum {{n^6}} } \right)}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{\dfrac{1}{{24}}\left( {2{n^7} + 4{n^6} + 2{n^5} + 3{n^6} + 6{n^5} + 3{n^4} + {n^5} + 2{n^4} + {n^3}} \right)}}{{\dfrac{1}{{42}}\left( {6{n^7} + 21{n^6} + 21{n^5} - 7{n^3} + n} \right)}}$
Taking the constant term $\left( {\dfrac{{\dfrac{1}{{24}}}}{{\dfrac{1}{{42}}}} = \dfrac{{42}}{{24}} = \dfrac{7}{4}} \right)$ out of the limit and taking ${n^7}$ common from both the numerator and the denominator, we get
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{{\left( {\sum {{n^2}} } \right)\left( {\sum {{n^3}} } \right)}}{{\left( {\sum {{n^6}} } \right)}} = \dfrac{7}{4}\mathop {\lim }\limits_{n \to \infty } \dfrac{{{n^7}\left( {2 + \dfrac{4}{n} + \dfrac{2}{{{n^2}}} + \dfrac{3}{n} + \dfrac{6}{{{n^2}}} + \dfrac{3}{{{n^3}}} + \dfrac{1}{{{n^2}}} + \dfrac{2}{{{n^3}}} + \dfrac{1}{{{n^4}}}} \right)}}{{{n^7}\left( {6 + \dfrac{{21}}{n} + \dfrac{{21}}{{{n^2}}} - \dfrac{7}{{{n^4}}} + \dfrac{1}{{{n^6}}}} \right)}}$
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{{\left( {\sum {{n^2}} } \right)\left( {\sum {{n^3}} } \right)}}{{\left( {\sum {{n^6}} } \right)}} = \dfrac{7}{4}\mathop {\lim }\limits_{n \to \infty } \dfrac{{\left( {2 + \dfrac{4}{n} + \dfrac{2}{{{n^2}}} + \dfrac{3}{n} + \dfrac{6}{{{n^2}}} + \dfrac{3}{{{n^3}}} + \dfrac{1}{{{n^2}}} + \dfrac{2}{{{n^3}}} + \dfrac{1}{{{n^4}}}} \right)}}{{\left( {6 + \dfrac{{21}}{n} + \dfrac{{21}}{{{n^2}}} - \dfrac{7}{{{n^4}}} + \dfrac{1}{{{n^6}}}} \right)}}$
Applying the limit $n \to \infty$ , we get
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{{\left( {\sum {{n^2}} } \right)\left( {\sum {{n^3}} } \right)}}{{\left( {\sum {{n^6}} } \right)}} = \dfrac{7}{4}\left( {\dfrac{2}{6}} \right) = \dfrac{7}{{12}}$ $\left[ {\because {\text{ as n}} \to \infty {\text{ then }}\dfrac{1}{n} \to \dfrac{1}{\infty } = 0} \right]$
Therefore, the value of $\mathop {\lim }\limits_{n \to \infty } \dfrac{{\left( {\sum {{n^2}} } \right)\left( {\sum {{n^3}} } \right)}}{{\left( {\sum {{n^6}} } \right)}}$ is $\dfrac{7}{{12}}$.
Hence, option (D) is correct.

Note: In such questions, you should know the formula of the sum of the series or expansion used since these are questions which have direct implementation of the standard formulae and are quite easy yet lengthy but you can score well. You can also solve this question by using L’ Hopital’s rule but it would be more time consuming and complicated as well.