How to find the value of the given limit with summation?

Question

It was asked to find the value of the following limit: $$ \lim _{n \rightarrow \infty} \frac{96}{n^4}\left[1\left(\sum_{k=1}^n k\right)+2\left(\sum_{k=1}^{n-1} k\right)+3\left(\sum_{k=1}^{n-2} k\right)+\cdots+n \cdot 1\right] $$ Here's my try on it:

Let the general $m^{th}$ term of the series be $$\begin{align}t_m &= m \left( \displaystyle \sum_{k=1}^{n-m+1}k \right)\\ &= m\ .\dfrac{(n-m+1)(n-m+2)}{2} \end{align}$$ So, $$ \begin{align} \sum_{n=1}^n t_m &=\sum_{m=1}^n \frac{m(n-m+1)(n-m+2)}{2} \\ &=\sum_{s=0}^{n-1} \frac{(s+1)(n-s)(n-s+1)}{2} \end{align} $$ From here I have no clue how to proceed further with this series.
Thanks for any help.

Answer 1

If we look at the inside sum we have $$ \sum_{k=1}^nk\cdot\sum_{i=1}^{n-k+1}i $$ Now we have that the inside sum is the triangle numbers namely we have that $$ \sum_{i=1}^{n-k+1}i=\frac{(n-k+1)(n-k+2)}{2} $$ Thus, we have that the sum becomes $$ \sum_{k=1}^n\frac{k(n-k+1)(n-k+2)}{2} $$ Now if we factor everything out and gather things in terms of $k$ we get $$ \frac{1}{2}\sum_{k=1}^nk^3-(2n+3)k^2+(n^2+3n+2)k $$ Then if we break up the sums we have $$ \frac{1}{2}\sum_{k=1}^nk^3 -\frac{2n+3}{2}\sum_{k=1}^nk^2+\frac{n^2+3n+2}{2}\sum_{k=1}^nk $$ Now if we use the well-known formulas $$ \sum_{k=1}^nk^3=\left(\frac{n(n+1)}{2}\right)^2 $$

$$ \sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6} $$ and the formula for the triangle numbers above, we then see that the sum becomes $$ \frac{1}{2}\left(\frac{n(n+1)}{2}\right)^2-\frac{2n+3}{2}\frac{n(n+1)(2n+1)}{6}+\frac{n^2+3n+2}{2}\frac{n(n+1)}{2} $$ Now we are really only interested in the $n^4$ term as the lower order terms will be killed when we take the limit.

If we see from the first term we will get a factor of $\frac{n^4}{8}+O(n^3)$, the second factor, we will get a factor of $-\frac{n^4}{3}+O(n^3)$, and the last term will give us a $\frac{n^4}{4}+O(n^3)$

Thus, we have the above is $$ \frac{1}{24}n^4+O(n^3) $$ Thus, when we consider $$ \lim_{n\rightarrow \infty}\frac{96}{n^4}\cdot\text{sum}=\lim_{n\rightarrow \infty}\frac{96}{n^4}\cdot\left(\frac{1}{24}n^4+O(n^3)\right)=4 $$

Answer 2

Let $A_n=\sum_{j=1}^{n}\left[j\sum_{k=1}^{n-j+1} k\right]$. Write out $A_{n-1}$ and notice that $A_n-A_{n-1}=\sum_{j=1}^{n}\sum_{k=1}^{j}k$ (write out terms by terms if necessary). Since $n^4$ is strictly monotone and divergent sequence, by Stolz–Cesàro

$$ \lim _{n \rightarrow \infty} \frac{A_n}{n^4}=\lim _{n \rightarrow \infty}\frac{A_n-A_{n-1}}{n^4-(n-1)^4}=\lim _{n \rightarrow \infty} \frac{\sum_{j=1}^{n}\sum_{k=1}^{j}k}{4n^3-6n^2+4n-1}. $$ Letting $B_n=\sum_{j=1}^{n}\sum_{k=1}^{j}k$ and applying Stolz–Cesàro similarly again

$$ \lim _{n \rightarrow \infty} \frac{B_n}{4n^3-6n^2+4n-1}=\lim _{n \rightarrow \infty} \frac{\sum_{k=1}^{n}k}{12n^2-24n+14}. $$ And just for consistency $C_n=\sum_{k=1}^{n}k$ and for the last time

$$ \lim _{n \rightarrow \infty} \frac{C_n}{12n^2-24n+14}=\lim_{n\to \infty}\frac{n}{24n-36}=\frac{1}{24}. $$ Hence the answer is $\frac{96}{24}=4$.

Answer 3

The given limit can be written as $$\begin{align}&\lim_{n\to\infty}\frac{96}{n^4}\sum_{m=1}^nm\sum_{k=1}^{n-m+1}k \\\\&=\lim_{n\to\infty}\frac{96}{n^4}\sum_{m=1}^{n}m\cdot\frac{(n-m+1)(n-m+2)}{2} \\\\&=\lim_{n\to\infty}\frac{48}{n}\sum^{n}_{m=1}\frac mn\bigg(1-\frac mn+\frac 1n\bigg)\bigg(1-\frac mn+\frac 2n\bigg) \\\\&=\lim_{n\to\infty}\frac{48}{n}\sum_{m=1}^{n}\frac mn\bigg(\bigg(1-\frac mn\bigg)^2+\frac 3n\bigg(1-\frac mn\bigg)+\frac{2}{n^2}\bigg) \\\\&=\lim_{n\to\infty}\frac{48}{n}\sum_{m=1}^{n}\frac mn\bigg(1-\frac mn\bigg)^2 \\&\qquad +\lim_{n\to\infty}\frac{48}{n}\sum_{m=1}^{n}\frac mn\bigg(\frac 3n\bigg(1-\frac mn\bigg)+\frac{2}{n^2}\bigg) \\\\&=\lim_{n\to\infty}\frac{48}{n}\sum_{m=1}^{n}\frac mn\bigg(1-\frac mn\bigg)^2 \\&\qquad +\lim_{n\to\infty}\frac{48}{n^4}\sum_{m=1}^{n}(3mn-3m^2+2m)\tag1\end{align}$$

The second term is equal to $0$, so we get $$\begin{align}(1)&=48\int_0^1x(1-x)^2dx \\\\&=48\int_0^1(x-2x^2+x^3)dx \\\\&=48\bigg(\frac 12-\frac 23+\frac 14\bigg) \\\\&=\color{red}4\end{align}$$

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Added :

Ivan Kaznacheyeu's comment is helpful.

First of all, we have $$\lim_{n\to\infty}\frac{96}{n^4}\sum_{m=1}^{n}m\sum_{k=1}^{n-m+1}k=\lim_{n\to\infty}\frac{96}{n^4}\sum_{m=1}^{n}m\sum_{k=1}^{\color{red}{n-m}}k\tag1$$ because $$\begin{align}&\lim_{n\to\infty}\frac{96}{n^4}\sum_{m=1}^{n}m\sum_{k=1}^{n-m+1}k \\\\&=\lim_{n\to\infty}\frac{96}{n^4}\sum_{m=1}^{n}m\bigg(n-m+1+\sum_{k=1}^{\color{red}{n-m}}k\bigg) \\\\&=\underbrace{\lim_{n\to\infty}\frac{96}{n^4}\sum_{m=1}^{n}m(n-m+1)}_{=0} \\&\qquad +\lim_{n\to\infty}\frac{96}{n^4}\sum_{m=1}^{n}m\sum_{k=1}^{\color{red}{n-m}}k\end{align}$$

Now, $(1)$ can be written as $$(1)=96\lim_{n\to\infty}\frac{1}{n}\sum_{m=1}^{n}\frac mn\underbrace{\bigg(\frac 1n\sum_{k=1}^{n-m}\frac kn\bigg)}_{S}\tag2$$

Here, $S$ is as follows :

Therefore, we finally get $$\begin{align}(2)&=96\int_{0}^{1}x\bigg(\int_{0}^{1-x}y\ dy\bigg)dx \\\\&=96\int_0^1\frac{x(1-x)^2}{2}dx \\\\&=4\end{align}$$