Question

The limit of $\dfrac{1}{{{n}^{4}}}\sum\nolimits_{k=1}^{n}{k\left( k+2 \right)\left( k+4
\right)}$as $n\to \infty $
(A) Exists and equals $\dfrac{1}{4}$
(B) Exists and equals $0$
(C) Exists and equals $\dfrac{1}{8}$
(D) Does not exists

Answer 1

Hint: Try thinking what the numerator would look like on simplification of the series given to us in terms
of $k$. Then, take the highest degree of $n$ common from the numerator, and perform further
simplification, before applying the limit, to arrive at the answer.
Firstly, we have to check it by putting the $\underset{n\to \infty }{\mathop{\lim }}\,$ and
substitute the value of $k$ accordingly.

$\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{{{n}^{4}}}\sum\nolimits_{k=1}^{n}{k\left( k+2
\right)\left( k+4 \right)}$
$=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{{{n}^{4}}}\left[ \left( 1\left( 1+2 \right)\left( 1+4
\right) \right)+\left( 2\left( 2+2 \right)\left( 2+4 \right) \right)+......+n\left( n+2 \right)\left( n+4 \right)
\right]$
$=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{{{n}^{4}}}\left[ 15+48+......+n\left( n+2 \right)\left(
n+4 \right) \right]$
$=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{a{{n}^{3}}+b{{n}^{2}}+cn+d}{{{n}^{4}}}$
Over here, whatever we have inside the bracket can be written in the general form of a cubic
polynomial, or a polynomial of degree 3, where the variable is $n$, and $a,b,c,d$ are constants that I’ve
assumed for the final polynomial that we will get if we actually simplify the expression inside the
bracket. Thus, we can conclude that the numerator is a polynomial of degree 3.
Now, taking ${{n}^{3}}$ common in the numerator, we’ll get :
$=\underset{n\to \infty }{\mathop{\lim
}}\,\dfrac{{{n}^{3}}(a+\dfrac{b}{n}+\dfrac{c}{{{n}^{2}}}+\dfrac{d}{{{n}^{3}}})}{{{n}^{4}}}$
Now, if we analyse the numerator, or the term inside the bracket, more specifically, we’ll see that the
terms contain some power of the term $\dfrac{1}{n}$ multiplied by the constant.
Now, if we put $\underset{n\to \infty }{\mathop{\lim }}\,$ for the term $\dfrac{1}{n}$, we’ll get
$\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}=\dfrac{1}{\infty }=0$.
Using this relation now, we can simplify the numerator further. Substituting for $\dfrac{1}{n}$, we’ll get
:
$=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{{{n}^{3}}(a)}{{{n}^{4}}}=\underset{n\to \infty
}{\mathop{\lim }}\,\dfrac{a}{n}$
Since $a$ is a constant here, we can simply now put the limiting value of $n$, which is $\infty $. Doing
so, we’ll get :
$\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{a}{n}=\dfrac{a}{\infty }=0$
Therefore, we can now see that the limit exists, and is equal to $0$.
Option (B) is the correct answer, exists and equal to $0$.
Note: Student make the mistakes by directly solving it, but solve it step by step before putting the limit
and put the limits into it. Remember, $\infty +\text{ }$any number $=\infty $ , since $\infty $ is not
defined, and can be any number that’s big enough.