Evaluate the summation of :-

1.	Evaluate the summation of :-
Answer» We're ASKED to EVALUATE the LIMIT, $\small\text{$\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, $\small\text{$\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 $\small\text{$\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, $\small\text{$\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]$}$ $\small\text{$\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]$}$ $\small\text{$\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]$}$ $\small\text{$\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, $\small\text{$\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, $\small\text{$\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]$}$ $\small\text{$\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 $\small\text{$\displaystyle\lim_{n\to\infty}\dfrac{1}{n}=0,$}$ $\small\text{$\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]$}$ $\small\text{$\displaystyle\longrightarrow L=\dfrac{1}{2}\left[\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{1}{4}\right]$}$ $\small\text{$\displaystyle\longrightarrow L=\dfrac{1}{2}\cdot\dfrac{1}{12}$}$ $\small\text{$\displaystyle\longrightarrow\underline{\underline{L=\dfrac{1}{24}}}$}$

Answer»

$\small\text{$\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,

$\small\text{$\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 $\small\text{$\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,

$\small\text{$\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]$}$

$\small\text{$\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]$}$

$\small\text{$\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]$}$

$\small\text{$\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,

$\small\text{$\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,

$\small\text{$\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]$}$

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

$\small\text{$\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]$}$