Question

Value of $L = \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{{n^4}}}\left[ {1 \cdot \left( {\sum\limits_{k = 1}^n k } \right) + 2 \cdot \left( {\sum\limits_{k = 1}^{n - 1} k } \right) + 3 \cdot \left( {\sum\limits_{k = 1}^{n - 2} k } \right) + ... + n \cdot 1} \right]$ is?
A.$\dfrac{1}{{24}}$
B.$\dfrac{1}{{12}}$
C.$\dfrac{1}{6}$
D.$\dfrac{1}{3}$

Answer 1

Here we are asked to find the value of the given sequence which has been given limit $n \to \infty$. To solve this problem, we first need to find the ${r^{th}}$ term (general form of any term). Then we will be using some standard summation formulae to expand the inner summation in the given problem. Then we will simplify it to find the required solution. Formulae that we use are given in the following formula section.
Formulae to be used:
$\sum\limits_{k = 1}^n k = \dfrac{{n(n + 1)}}{2}$
$\sum\limits_{k = 1}^n {{k^2}} = \dfrac{{n(n + 1)(2n + 1)}}{6}$
$\sum\limits_{k = 1}^n {{k^3}} = {\left( {\dfrac{{n(n + 1)}}{2}} \right)^2}$
$\mathop {\lim }\limits_{x \to \infty } \dfrac{1}{x} = 0$

Complete step by step answer:
Given sequence $L = \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{{n^4}}}\left[ {1 \cdot \left( {\sum\limits_{k = 1}^n k } \right) + 2 \cdot \left( {\sum\limits_{k = 1}^{n - 1} k } \right) + 3 \cdot \left( {\sum\limits_{k = 1}^{n - 2} k } \right) + ... + n \cdot 1} \right]$ .
To find the limit of the given sequence when $n \to \infty$ .
First, we know, $\sum\limits_{k = 1}^n k = \dfrac{{n(n + 1)}}{2}$ , then, the second term becomes, $2\sum\limits_{k = 1}^{n - 1} k = \dfrac{{2(n - 1)(n)}}{2}$ , similarly, the third term becomes, $3\sum\limits_{k = 1}^{n - 2} k = \dfrac{{3(n - 2)(n - 1)}}{2}$ .
In similar manner, the ${r^{th}}$ term becomes, ${T_r} = \dfrac{{r(n + 1 - r)(n + 2 - r)}}{2}$ .
Now, consider $(n + 1)$ as one term and similarly, $(n + 2)$ as one term and multiply both brackets and simplify,
${T_r} = \dfrac{{r\left[ {(n + 1)(n + 2) - r(n + 1) - r(n + 2) + {r^2}} \right]}}{2}$ which can also be written as ${T_r} = \dfrac{{r\left[ {(n + 1)(n + 2) - r\left( {(n + 1) + (n + 2)} \right) + {r^2}} \right]}}{2}$ .
Now multiplying $r$ inside the brackets, we get ${T_r} = \dfrac{{\left[ {r(n + 1)(n + 2) - {r^2}(2n + 3) + {r^3}} \right]}}{2}$ , which can also be written as ${T_r} = \dfrac{{r(n + 1)(n + 2)}}{2} - \dfrac{{{r^2}(2n + 3)}}{2} + \dfrac{{{r^3}}}{2}$ .
Applying summation on both sides, we get, $\sum\limits_{r = 1}^n {{T_r}} = \sum\limits_{r = 1}^n {\dfrac{{r(n + 1)(n + 2)}}{2}} - \sum\limits_{r = 1}^n {\dfrac{{{r^2}(2n + 3)}}{2}} + \sum\limits_{r = 1}^n {\dfrac{{{r^3}}}{2}}$ .
Now, using formulae associated with summation, we get, $\sum\limits_{r = 1}^n {{T_r}} = \dfrac{{n(n + 1)(n + 1)(n + 2)}}{4} - \dfrac{{n(n + 1)(2n + 1)(2n + 3)}}{{12}} + \dfrac{{{{\left( {n(n + 1)} \right)}^2}}}{8}$ .
Now, take ${n^4}$ common from each numerator, we get, $\sum\limits_{r = 1}^n {{T_r}} = {n^4}\left[ {\dfrac{{1 \cdot \left( {1 + \dfrac{1}{n}} \right)\left( {1 + \dfrac{1}{n}} \right)\left( {1 + \dfrac{2}{n}} \right)}}{4} - \dfrac{{1 \cdot \left( {1 + \dfrac{1}{n}} \right)\left( {2 + \dfrac{1}{n}} \right)\left( {2 + \dfrac{3}{n}} \right)}}{{12}} + \dfrac{{{{\left( {1 \cdot \left( {1 + \dfrac{1}{n}} \right)} \right)}^2}}}{8}} \right]$ .
Now, finally applying limits, we get, $\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{{n^4}}}\sum\limits_{r = 1}^n {{T_r}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{{n^4}}}{{{n^4}}}\left[ {\dfrac{{1 \cdot \left( {1 + \dfrac{1}{n}} \right)\left( {1 + \dfrac{1}{n}} \right)\left( {1 + \dfrac{2}{n}} \right)}}{4} - \dfrac{{1 \cdot \left( {1 + \dfrac{1}{n}} \right)\left( {2 + \dfrac{1}{n}} \right)\left( {2 + \dfrac{3}{n}} \right)}}{{12}} + \dfrac{{{{\left( {1 \cdot \left( {1 + \dfrac{1}{n}} \right)} \right)}^2}}}{8}} \right]$
We know, $\dfrac{1}{n} \to 0$ when $n \to \infty$ , so we get, $\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{{n^4}}}\sum\limits_{r = 1}^n {{T_r}} = \mathop {\lim }\limits_{n \to \infty } \left[ {\dfrac{{1 \cdot \left( {1 + 0} \right)\left( {1 + 0} \right)\left( {1 + 0} \right)}}{4} - \dfrac{{1 \cdot \left( {1 + 0} \right)\left( {2 + 0} \right)\left( {2 + 0} \right)}}{{12}} + \dfrac{{{{\left( {1 \cdot \left( {1 + 0} \right)} \right)}^2}}}{8}} \right]$ i.e., $\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{{n^4}}}\sum\limits_{r = 1}^n {{T_r}} = \left[ {\dfrac{1}{4} - \dfrac{4}{{12}} + \dfrac{1}{8}} \right]$ , now, least common multiple of $4,12,8$ is $24$ .
Hence, we get, $\dfrac{{6 - 8 + 3}}{{24}}$ i.e., $\dfrac{1}{{24}}$ .

Thus, option $(A)$ is correct.

Note:
One can easily verify if the ${r^{th}}$ term is correct or not by putting different values of $r$ . Like for this question, the ${r^{th}}$ term is ${T_r} = \dfrac{{r(n + 1 - r)(n + 2 - r)}}{2}$ , if we put $r = 1$ , then ${T_1} = \dfrac{{n(n + 1)}}{2}$ , which matches our first term, similarly, for $r = 2$ , ${T_2} = \dfrac{{2(n + 1 - 2)(n + 2 - 2)}}{2} = \dfrac{{2(n - 1)(n)}}{2}$ and finally if we put $r = n$ , then ${T_n} = \dfrac{{n(n + 1 - n)(n + 2 - n)}}{2} = \dfrac{{n(1)(2)}}{2} = n \cdot 1$ , which is our ${n^{th}}$ term.
In limit $n \to \infty$ , $n$ is not exactly equal to $\infty$ , but it is approaching towards $\infty$ , or we can say that $n$ is tending towards $\infty$ .
One should remember the basic identities associated with summation, which makes our work easier in solving such questions.