Question

Evaluate the summation of :-

$\displaystyle \rm\footnotesize{ \lim\limits_{n\to\infty} \dfrac1{n^4}\left[1\bigg(\sum\limits_{k=1}^nk\bigg)+2\bigg(\sum\limits_{k=1}^{n-1}k\bigg)+3\bigg(\sum\limits_{k=1}^{n-2}k\bigg)+...+n.1\right]}$
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Answer 1

We're asked to evaluate the limit,

$\small\displaystyle\longrightarrow L=\lim_{n\to\infty}\dfrac{1}{n^4}\left[1\left(\sum_{k=1}^nk\right)+2\left(\sum_{k=1}^{n-1}k\right)+3\left(\sum_{k=1}^{n-2}k\right)+\,\dots\,+n\cdot1\right]$

or,

$\small\displaystyle\longrightarrow L=\lim_{n\to\infty}\dfrac{1}{n^4}\left[\sum_{r=1}^nr\left(\sum_{k=1}^{n-(r-1)}k\right)\right]=\lim_{n\to\infty}\dfrac{1}{n^4}\left[\sum_{r=1}^nr\left(\sum_{k=1}^{n-r+1}k\right)\right]$

Since $\small\displaystyle\sum_{k=1}^mk=\dfrac{m(m+1)}{2},\ \sum_{k=1}^{n-r+1}k=\dfrac{(n-r+1)(n-r+2)}{2}.$ Then,

$\small\displaystyle\longrightarrow L=\lim_{n\to\infty}\dfrac{1}{n^4}\left[\sum_{r=1}^nr\cdot\dfrac{(n-r+1)(n-r+2)}{2}\right]$

$\small\displaystyle\longrightarrow L=\dfrac{1}{2}\lim_{n\to\infty}\dfrac{1}{n^4}\left[\sum_{r=1}^nr(n+1-r)(n+2-r)\right]$

$\small\displaystyle\longrightarrow L=\dfrac{1}{2}\lim_{n\to\infty}\dfrac{1}{n^4}\left[\sum_{r=1}^n[r(n+1)(n+2)-r^2(2n+3)+r^3]\right]$

$\small\displaystyle\longrightarrow L=\dfrac{1}{2}\lim_{n\to\infty}\dfrac{1}{n^4}\left[(n+1)(n+2)\sum_{r=1}^nr-(2n+3)\sum_{r=1}^nr^2+\sum_{r=1}^nr^3\right]$

We have,

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

Then,

$\small\displaystyle\longrightarrow L=\dfrac{1}{2}\lim_{n\to\infty}\dfrac{1}{n^4}\left[(n+1)(n+2)\cdot\dfrac{n(n+1)}{2}-(2n+3)\cdot\dfrac{n(n+1)(2n+1)}{6}+\dfrac{n^2(n+1)^2}{4}\right]$

$\small\displaystyle\longrightarrow L=\dfrac{1}{2}\lim_{n\to\infty}\left[\dfrac{n(n+1)^2(n+2)}{2n^4}-\dfrac{n(n+1)(2n+1)(2n+3)}{6n^4}+\dfrac{n^2(n+1)^2}{4n^4}\right]$

$\small\text{ \displaystyle\longrightarrow L=\dfrac{1}{2}\lim_{n\to\infty}\left[\dfrac{\left(1+\frac{1}{n}\right)^2\left(1+\frac{2}{n}\right)}{2}-\dfrac{\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)\left(2+\frac{3}{n}\right)}{6}+\dfrac{\left(1+\frac{1}{n}\right)^2}{4}\right] }$

Since $\small\displaystyle\lim_{n\to\infty}\dfrac{1}{n}=0,$

$\small\displaystyle\longrightarrow L=\dfrac{1}{2}\left[\dfrac{(1+0)^2(1+2\cdot0)}{2}-\dfrac{(1+0)(2+0)(2+3\cdot0)}{6}+\dfrac{(1+0)^2}{4}\right]$

$\small\displaystyle\longrightarrow L=\dfrac{1}{2}\left[\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{1}{4}\right]$

$\small\displaystyle\longrightarrow L=\dfrac{1}{2}\cdot\dfrac{1}{12}$

$\small\text{ \displaystyle\longrightarrow\underline{\underline{L=\dfrac{1}{24}}} }$