Show that $\lim_{n\to\infty}\frac{1}{n^4}\sum_{j=1}^{n}\left((2j-1)\sum_{k=1}^{n+1-j}k\right)=\frac{1}{12}$ Show that$$\lim_{n\rightarrow \infty} \frac{1}{n^4} \left(1\left(\sum_{k=1}^{n}k\right)+ 3\left(\sum_{k=1}^{n-1}k\right)+5\left(\sum_{k=1}^{n-2}k\right)+\cdots+(2n-1)\cdot1\right)=\frac{1}{12}$$ 
This is getting more and more complicated as number of terms are decreasing.
 A: Hint: We have 
\begin{align*}
\sum_{j=1}^{n}\left((2j-1)\sum_{k=1}^{n+1-j}k\right)&=\sum_{j=1}^n\frac{(2j-1)(n+2-j)(n+1-j)}{2}\\
&=\frac{1}{2}\sum_{j=1}^n(-2 + 7 j - 7 j^2 + 2 j^3 - 3 n + 8 j n - 4 j^2 n - n^2 + 2 j n^2).
\end{align*}
In the limit, the only summands which will matter correspond to the $2j^3$, $4j^2n$, and the $2jn^2$, as all others will result in a polynomial in $n$ of degree less than $3$. Thus we have 
$$
\lim_{n\to\infty}\frac{1}{n^4}\sum_{j=1}^{n}\left(2j-1\sum_{k=1}^{n+1-j}k\right)=\lim_{n\to\infty}\frac{1}{2n^4}\sum_{j=1}^n(2j^3-4j^2n+2jn^2)
$$
Now try to use formulas for the sum of third, second, and first powers, respectively.
A: Hint:
You need to compute
$$\lim_{n\to \infty} \frac{1}{n^4} \sum_{a=0}^{n-1} \frac{(2a+1) (n-a)(n+1-a)}{2}=\lim_{n \to \infty}\frac{n (n+1)^2(n+2)}{12 n^4}.$$
A: $$\begin{align}
\sum_{r=1}^n\left[(2r-1)\sum_{k=1}^{n-r+1}k\right]
&=\sum_{r=1}^n\left[(2r-1)\sum_{s=r}^n (n-s+1)\right]
&&(s=n-k+1)\\
&=\sum_{r=1}^n\sum_{s=r}^n\sum_{t=s}^n(2r-1)\\
&=\sum_{t=1}^n\sum_{s=1}^t\sum_{r=1}^s (2r-1)
&&(1\le r\le s\le t\le n)\\
&=\sum_{t=1}^n \sum_{s=1}^t s^2\\
&=\sum_{t=1}^n \sum_{s=1}^t \binom s2+\binom {s+1}2\\
&=\sum_{t=1}^n \binom {t+1}3+\binom {t+2}3\\
&=\binom {n+2}4+\binom {n+3}4\\
&=\frac 14\big[(n-1)+(n+3)\big]\binom {n+2}3\\
&=\frac 12(n+1)\binom {n+2}3\\
&=\frac {(n+2)(n+1)^2n}{12}\\
\lim_{n\to \infty}\frac 1{n^4}\sum_{r=1}^n\left[(2r-1)\sum_{k=1}^{n-r+1}k\right]
&=\lim_{n\to\infty}\frac 1{n^4}\cdot\frac {(n+2)(n+1)^2n}{12}\\
&=\lim_{n\to\infty}\frac 1{n^4}\cdot \frac {n^4+O(n^3)}{12}\\
&=\lim_{n\to\infty}\frac {1+O(\frac 1n)}{12}\\
&=\frac 1{12}
\end{align}$$
A: You can convert the whole limit as $$\int_0^1 x(1-x)^2 dx$$ by using the limit of sum.
$$\lim_{n \to \infty} \frac1{n^4} \sum_{j=1}^n ((2j-1)\frac{(n+1-j)(n+2-j)}{2}) = \frac12 \lim_{n \to \infty} \frac1n \sum_{j=1}^{n} ((2 \frac{j}{n} -\frac1n)(1+\frac1n-\frac{j}{n})(1+\frac2n - \frac{j}{n}))$$ 
Now use the formula for limit of a sum
