Question

The value of $$\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{\left(\sum n^2\right)\left(\sum n^3\right)}{\left(\sum n^6\right)}}$$ is
(a)1
(b)$${\displaystyle \frac{3}{2}}$$
(c)$${\displaystyle \frac{5}{6}}$$
(d)$${\displaystyle \frac{7}{12}}$$

Answer 1

Hint: Here, we will first use the formulae of sums, $$\sum n^2={\displaystyle \frac{n\left(n+1\right)\left(n+2\right)}{6}}$$, $$\sum n^3={\left[{\displaystyle \frac{n\left(n+1\right)}{2}}\right]}^2$$ and $$\sum n^6={\displaystyle \frac{1}{42}}\left(6n^7+21n^6+21n^5-7n^3+n\right)$$ in the given expression and then taking $$n\to \mathrm{\infty }$$ on right hand side of the above equation, $${\displaystyle \frac{1}{n}}\to 0$$ to find the required value.

Complete step-by-step answer:
We are given $$\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{\left(\sum n^2\right)\left(\sum n^3\right)}{\left(\sum n^6\right)}}$$.

Using the formulae of sums, $$\sum n^2={\displaystyle \frac{n\left(n+1\right)\left(n+2\right)}{6}}$$, $$\sum n^3={\left[{\displaystyle \frac{n\left(n+1\right)}{2}}\right]}^2$$ and $$\sum n^6={\displaystyle \frac{1}{42}}\left(6n^7+21n^6+21n^5-7n^3+n\right)$$ in the above expression, we get


$$\begin{gathered} \Rightarrow \underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{\left(\sum n^2\right)\left(\sum n^3\right)}{\left(\sum n^6\right)}}=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{\left[{\displaystyle \frac{n\left(n+1\right)\left(n+2\right)}{6}}\right]{\left[{\displaystyle \frac{n\left(n+1\right)}{2}}\right]}^2}{\left[{\displaystyle \frac{6n^7+21n^6+21n^5-7n^3+n}{42}}\right]}} \\ \Rightarrow \underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{\left(\sum n^2\right)\left(\sum n^3\right)}{\left(\sum n^6\right)}}=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{\left[{\displaystyle \frac{2n^3+3n^2+n}{6}}\right]\left[{\displaystyle \frac{n^4+2n^3+n^2}{4}}\right]}{\left[{\displaystyle \frac{6n^7+21n^6+21n^5-7n^3+n}{42}}\right]}} \\ \Rightarrow \underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{\left(\sum n^2\right)\left(\sum n^3\right)}{\left(\sum n^6\right)}}={\displaystyle \frac{\left[{\displaystyle \frac{\left(2n^3+3n^2+n\right)\left(n^4+2n^3+n^2\right)}{24}}\right]}{\left[{\displaystyle \frac{6n^7+21n^6+21n^5-7n^3+n}{42}}\right]}} \\ \Rightarrow \underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{\left(\sum n^2\right)\left(\sum n^3\right)}{\left(\sum n^6\right)}}=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{42\left[2n^7+4n^6+3n^6+6n^5+3n^4+n^5+2n^4+n^3\right]}{24\left[6n^7+21n^6+21n^5-7n^3+n\right]}} \\ \Rightarrow \underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{\left(\sum n^2\right)\left(\sum n^3\right)}{\left(\sum n^6\right)}}=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{7\left[2n^7+4n^6+3n^6+6n^5+3n^4+n^5+2n^4+n^3\right]}{4\left[6n^7+21n^6+21n^5-7n^3+n\right]}} \end{gathered}$$


Dividing the numerator and denominator by $$n^7$$ in right side of the above equation, we get


$$\begin{gathered} \Rightarrow \underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{\left(\sum n^2\right)\left(\sum n^3\right)}{\left(\sum n^6\right)}}=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{7\left[{\displaystyle \frac{2n^7+4n^6+3n^6+6n^5+3n^4+n^5+2n^4+n^3}{n^7}}\right]}{4\left[{\displaystyle \frac{6n^7+21n^6+21n^5-7n^3+n}{n^7}}\right]}} \\ \Rightarrow \underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{\left(\sum n^2\right)\left(\sum n^3\right)}{\left(\sum n^6\right)}}=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{7\left[{\displaystyle \frac{2n^7}{n^7}}+{\displaystyle \frac{4n^6}{n^7}}+{\displaystyle \frac{2n^5}{n^7}}+{\displaystyle \frac{3n^6}{n^7}}+{\displaystyle \frac{6n^5}{n^7}}+{\displaystyle \frac{3n^4}{n^7}}+{\displaystyle \frac{n^5}{n^7}}+{\displaystyle \frac{2n^4}{n^7}}+{\displaystyle \frac{n^3}{n^7}}\right]}{4\left[{\displaystyle \frac{6n^7}{n^7}}+{\displaystyle \frac{21n^6}{n^7}}+{\displaystyle \frac{21n^5}{n^7}}-{\displaystyle \frac{7n^3}{n^7}}+{\displaystyle \frac{n}{n^7}}\right]}} \\ \Rightarrow \underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{\left(\sum n^2\right)\left(\sum n^3\right)}{\left(\sum n^6\right)}}=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{7\left[2+{\displaystyle \frac{4}{n}}+{\displaystyle \frac{2}{n^2}}+{\displaystyle \frac{3}{n}}+{\displaystyle \frac{6}{n^2}}+{\displaystyle \frac{3}{n^3}}+{\displaystyle \frac{1}{n^2}}+{\displaystyle \frac{2}{n^3}}+{\displaystyle \frac{1}{n^4}}\right]}{4\left[6+{\displaystyle \frac{21}{n}}+{\displaystyle \frac{21}{n^2}}-{\displaystyle \frac{7}{n^4}}+{\displaystyle \frac{1}{n^6}}\right]}} \end{gathered}$$

When taking $$n\to \mathrm{\infty }$$ on right hand side of the above equation, $${\displaystyle \frac{1}{n}}\to 0$$, we get

$$\begin{gathered} \Rightarrow \underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{\left(\sum n^2\right)\left(\sum n^3\right)}{\left(\sum n^6\right)}}={\displaystyle \frac{7\left[2+0+0+0+0+0+0+0+0\right]}{4\left[6+0+0-0+0\right]}} \\ \Rightarrow \underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{\left(\sum n^2\right)\left(\sum n^3\right)}{\left(\sum n^6\right)}}={\displaystyle \frac{7\cdot 2}{4\cdot 6}} \\ \Rightarrow \underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{\left(\sum n^2\right)\left(\sum n^3\right)}{\left(\sum n^6\right)}}={\displaystyle \frac{7}{12}} \end{gathered}$$


Note: Whenever we face such types of questions on summation problems, students must remember the basic summation formulae of the series. Students must not get confused with the values of sums, as the main part of the question will be over then.