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Find the value of $\displaystyle\lim_{n \rightarrow \infty} \dfrac 1{n^4}\left[1\left(\sum^n_{k=1}k\right)+2\left(\sum^{n-1}_{k=1}k\right)+3\left(\sum^{n-2}_{k=1}k\right)+\dots+n.1\right]$

Please guide how to proceed in this case .....Thanks..

Here answer is :

The required limit is $\dfrac{1}{24}$

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Sultan, you have three great answers, I think you should accept one of them. :) –  Lays Apr 22 '13 at 3:55
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3 Answers

The sum of the first $n$ natural numbers is well-known to be $$ \sum_{k=1}^nk=\frac{n(n+1)}{2}. $$ Thus the expression within the brackets can be written as $$ \left[\frac{n(n+1)}{2}+2\frac{(n-1)n}{2}+3\frac{(n-2)(n-1)}{2}+\cdots +n\right] $$ or in sum-notation as $$ \sum_{k=1}^n\left(k\frac{(n+1-k)(n+2-k)}{2}\right). $$ Now multiply out the parenthesis in the numerator and sum each term seperately - this only requires knowledge of the sums: $$ \sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6},\qquad \sum_{k=1}^n k^3=\frac{n^2(n+1)^2}{4}. $$ Finally, divide each term by $n^4$ and let $n\to\infty$.

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$$\displaystyle\lim_{n \rightarrow \infty} \dfrac 1{n^4}\left[1\left(\sum^n_{k=1}k\right)+2\left(\sum^{n-1}_{k=1}k\right)+3\left(\sum^{n-2}_{k=1}k\right)+\dots+n.1\right]=$$ $$\lim\limits_{n\rightarrow\infty}\dfrac{1}{n^4}\left[1\dfrac{n(n+1)}{2}+2\dfrac{(n-1)(n)}{2}+\dots +k\dfrac{(n-k+1)(n-k+2)}{2}+\dots+n\cdot1\right]=$$ $$\lim\limits_{n\rightarrow\infty}\dfrac{1}{n^4}\sum\limits_{k=1}^{n}k\dfrac{(n-k+1)(n-k+2)}{2}=$$ $$\lim\limits_{n\rightarrow\infty}\dfrac{1}{n^4}\sum\limits_{k=1}^{n}\dfrac{k}{2}((n-k)^2+3(n-k)+2)=$$ $$\lim\limits_{n\rightarrow\infty}\dfrac{1}{n^4}\left(\sum\limits_{k=1}^{n}\dfrac{k}{2}\left((n-k)^2+3(n-k)\right)+\dfrac{n(n+1)}{2}\right)=$$ $$\lim\limits_{n\rightarrow\infty}\dfrac{1}{n^4}\sum\limits_{k=1}^\infty\dfrac{k(n-k)^2}{2}+\lim\limits_{n\rightarrow\infty}\dfrac{1}{n^4}\sum\limits_{k=1}^\infty\dfrac{3k(n-k)}{2}+\lim\limits_{n\rightarrow\infty}\dfrac{1}{n^4}\dfrac{n(n+1)}{2}=$$ $$\lim\limits_{n\rightarrow\infty}\dfrac{1-0}{n}\sum\limits_{k=1}^\infty\dfrac{\frac{k}{n}(1-\frac{k}{n})^2}{2}+\lim\limits_{n\rightarrow\infty}\dfrac{1-0}{n^2}\sum\limits_{k=1}^\infty\dfrac{3\frac{k}{n}(1-\frac{k}{n})}{2}+0=$$ $$\int_{0}^{1}\dfrac{x(1-x)^2}{2}dx+0\cdot \int_{0}^{1}\dfrac{3x(1-x)}{2}dx=\dfrac{1}{24}$$

Note: We used the fact that the limit of Riemann Sums converges to the integral.

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You can write your expression as

$$\frac{1}{n^4}\sum_{l=1}^{n}l\sum_{k=1}^{n+1-l}k= \frac{1}{n^4}\sum_{l=1}^{n}l\frac{(n+1-l)(n+2-l)}{2} = \\ = \frac{1}{2n^4}\sum_{l=1}^{n}l(n^2+3n-2nl+2-3l +l^2)=\\ =\frac{1}{2n^4}\left\{ n^2\sum_{l=1}^{n}l+ 3n\sum_{l=1}^{n}l- 2n\sum_{l=1}^{n}l^2 +2\sum_{l=1}^{n}l-3\sum_{l=1}^{n}l^2+ \sum_{l=1}^{n}l^3 \right\}$$

Now consider that only those terms with $n^4$ will have a non-zero limit (all sums with powers $l^p$ will contribute a term with $n^{p+1}$). All lower order terms have a limit of zero. This means the limit can be written as

$$\lim_{n\rightarrow\infty}\frac{1}{2n^4}\left\{\frac{n^4}{2} -\frac{2n^4}{3} + \frac{n^4}{4}\right\}= \lim_{n\rightarrow\infty}\frac{n^4}{24n^4} = \frac{1}{24} $$

[Note that you don't need to remember (or look up) all formulas for the sums $\sum_{l=1}^{n}l^p$, as long as you know that $\sum_{l=1}^{n}l^p = \frac{n^{p+1}}{p+1} +$ lower order terms.]

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