# 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

Last updated date: 30th Mar 2023

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Answer

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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.

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.

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