Question

Evaluate the following limit :
$\displaystyle \sf \lim_{n \longrightarrow \: \infty } \: \dfrac{1 \times n + {2}^{2}(n - 1) + {3}^{2} (n - 2) + \cdots + {n}^{2} \times 1}{ {n}^{4} }$
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Answer 1

First the given series (in the numerator) is brought to be written under summation notation by its general term.

Here the general form of each term in the series is,

$\displaystyle\sf{\longrightarrow a_r=r^2(n-r+1)}$

So the sum can be written as summation of each $\displaystyle\sf{a_r=r^2(n-r+1)}$ where $\displaystyle\sf {r}$ ranges from $\displaystyle\sf {1}$ to $\displaystyle\sf {n.}$

Hence the whole limit will be,

$\displaystyle\sf{\longrightarrow L=\lim_{n\to\infty}\sum_{r=1}^n\dfrac{r^2(n-r+1)}{n^4}}$

That $\displaystyle\sf{\dfrac{1}{n^4}}$ can be taken outside the sum as its constant wrt $\displaystyle\sf {r.}$

Now $\displaystyle\sf {r^2(n-r+1)}$ can be expanded as $\displaystyle\sf {nr^2-r^3+r^2}$ and the sum can be distributed to each of the three terms.

After further simplifications, the following formulas are recalled.

• $\displaystyle\sf{\sum_{r=1}^nr^2=\dfrac{n(n+1)(2n+1)}{6}}$
• $\displaystyle\sf{\sum_{r=1}^nr^3=\dfrac{n^2(n+1)^2}{4}}$

On applying these sums and after further simplifications, we are getting three limits,

$\displaystyle\sf{\longrightarrow L_1=\lim_{n\to\infty}\dfrac{(n+1)(2n+1)}{n^2}}$

$\displaystyle\sf{\longrightarrow L_2=\lim_{n\to\infty}\left(\dfrac{n+1}{n}\right)^2}$

$\displaystyle\sf{\longrightarrow L_3=\lim_{n\to\infty}\dfrac{(n+1)(2n+1)}{n^3}}$

In each limit we make use of,

• $\displaystyle\sf {\lim_{n\to\infty}\dfrac{1}{n}=0}$

by rewriting them as in the figure.

Hence we get,

$\displaystyle\sf{\longrightarrow L_1=2}$

$\displaystyle\sf{\longrightarrow L_2=1}$

$\displaystyle\sf{\longrightarrow L_3=0}$

Finally we get $\displaystyle\bf {\dfrac{1}{12}}$ as the answer.

Answer 2

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

$(1). \: \boxed{ \pink{ \rm \: \sum_{r=1}^n \: r \: = \dfrac{n(n + 1)}{2} }}$

$(2). \: \boxed{ \pink{ \rm \: \sum_{r=1}^n} \: \pink{{r}^{2} = \dfrac{n(n + 1)(2n + 1)}{6} }}$

$(3). \: \boxed{ \pink{ \rm \: \sum_{r=1}^n \: {r}^{3} = \dfrac{ {n}^{2} {(n + 1)}^{2} }{4} }}$

$(4). \: \boxed{ \pink{ \rm \:\lim_{n\to\infty} \: \dfrac{1}{n} = 0 }}$

$\large\underline\purple{\bold{Solution :- }}$

Consider,

$\rm \lim_{n \longrightarrow \: \infty } \: \dfrac{1 \times n + {2}^{2}(n - 1) + {3}^{2} (n - 2) + \cdots + {n}^{2} \times 1}{ {n}^{4} }$

$\displaystyle \sf \lim_{n \longrightarrow \: \infty } \: \dfrac{1 \times (n - 1 + 1)+ {2}^{2}(n - 2 + 1) + {3}^{2} (n - 3 + 1) + \cdots + {n}^{2} \times (n - n + 1)}{ {n}^{4} }$

$\displaystyle \sf \lim_{n \longrightarrow \: \infty } \: \dfrac{ {1}^{2} \times (n - (1 - 1)+ {2}^{2}(n - (2 - 1)) + {3}^{2} (n - (3 - 1)) + \cdots + {n}^{2} \times (1 + (n - n)}{ {n}^{4} }$

$\rm \: \lim_{n\to\infty}\dfrac{\sum_{r=1}^n \: {r}^{2}(n - (r - 1)) }{ {n}^{4} }$

$\rm \: = \lim_{n\to\infty}\dfrac{\sum_{r=1}^n \: {r}^{2} (n - r + 1)}{ {n}^{4} }$

$\rm \: = \lim_{n\to\infty}\dfrac{\sum_{r=1}^n \: ( n{r}^{2} - {r}^{3} + {r}^{2} )}{ {n}^{4} }$

$= \rm \: \lim_{n\to\infty}\dfrac{n\sum_{r=1}^n {r}^{2} \: - \sum_{r=1}^n {r}^{3} + \sum_{r=1}^n {r}^{2} }{ {n}^{4} }$

$= \rm \: \lim_{n\to\infty}\dfrac{ {n}^{2} (n + 1)(2n + 1)}{ {6n}^{4} } - \lim_{n\to\infty}\dfrac{ {n}^{2} {(n + 1)}^{2} }{4 {n}^{4} } + \lim_{n\to\infty}\dfrac{n(n + 1)(2n + 1)}{6 {n}^{4} }$

$= \rm \: \lim_{n\to\infty}\dfrac{(1 + \dfrac{1}{n} )(2 + \dfrac{1}{n}) }{6} - \lim_{n\to\infty}\dfrac{ {(1 + \dfrac{1}{n} )}^{2} }{4} + \lim_{n\to\infty}\dfrac{(1 + \dfrac{1}{n} )(2 + \dfrac{1}{n} )}{6n}$

$\rm : \implies \:\dfrac{2}{6} - \dfrac{1}{4}$

$\rm : \implies \:\dfrac{1}{3} - \dfrac{1}{4}$

$\rm : \implies \:\dfrac{4 - 3}{12}$

$\rm : \implies \:\dfrac{1}{12}$