How to find limit of this function- with summation notation i.e Sigma notation 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 $\dfrac{1}{24}$
 A: 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 $n\to\infty$.
A: $$\displaystyle\lim_{n \rightarrow \infty} \dfrac 1{n^4}\left[1\left(\sum^n_{k=1}k\right)+2\left(\sum^{n-1}_{k=1}k\right)+3\left(\sum^{n-2}_{k=1}k\right)+\dots+n.1\right]=$$
$$\lim\limits_{n\rightarrow\infty}\dfrac{1}{n^4}\left[1\dfrac{n(n+1)}{2}+2\dfrac{(n-1)(n)}{2}+\dots +k\dfrac{(n-k+1)(n-k+2)}{2}+\dots+n\cdot1\right]=$$
$$\lim\limits_{n\rightarrow\infty}\dfrac{1}{n^4}\sum\limits_{k=1}^{n}k\dfrac{(n-k+1)(n-k+2)}{2}=$$
$$\lim\limits_{n\rightarrow\infty}\dfrac{1}{n^4}\sum\limits_{k=1}^{n}\dfrac{k}{2}((n-k)^2+3(n-k)+2)=$$
$$\lim\limits_{n\rightarrow\infty}\dfrac{1}{n^4}\left(\sum\limits_{k=1}^{n}\dfrac{k}{2}\left((n-k)^2+3(n-k)\right)+\dfrac{n(n+1)}{2}\right)=$$
$$\lim\limits_{n\rightarrow\infty}\dfrac{1}{n^4}\sum\limits_{k=1}^\infty\dfrac{k(n-k)^2}{2}+\lim\limits_{n\rightarrow\infty}\dfrac{1}{n^4}\sum\limits_{k=1}^\infty\dfrac{3k(n-k)}{2}+\lim\limits_{n\rightarrow\infty}\dfrac{1}{n^4}\dfrac{n(n+1)}{2}=$$
$$\lim\limits_{n\rightarrow\infty}\dfrac{1-0}{n}\sum\limits_{k=1}^\infty\dfrac{\frac{k}{n}(1-\frac{k}{n})^2}{2}+\lim\limits_{n\rightarrow\infty}\dfrac{1-0}{n^2}\sum\limits_{k=1}^\infty\dfrac{3\frac{k}{n}(1-\frac{k}{n})}{2}+0=$$
$$\int_{0}^{1}\dfrac{x(1-x)^2}{2}dx+0\cdot \int_{0}^{1}\dfrac{3x(1-x)}{2}dx=\dfrac{1}{24}$$
Note: We used the fact that the limit of Riemann Sums converges to the integral. 
A: 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.]
