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.

$$\begin{align}
\sum_{r=1}^n\left[(2r-1)\sum_{k=1}^{n-r+1}k\right]
&=\sum_{r=1}^n\left[(2r-1)\sum_{s=r}^n (n-s+1)\right]
&&(s=n-k+1)\\
&=\sum_{r=1}^n\sum_{s=r}^n\sum_{t=s}^n(2r-1)\\
&=\sum_{t=1}^n\sum_{s=1}^t\sum_{r=1}^s (2r-1)
&&(1\le r\le s\le t\le n)\\
&=\sum_{t=1}^n \sum_{s=1}^t s^2\\
&=\sum_{t=1}^n \sum_{s=1}^t \binom s2+\binom {s+1}2\\
&=\sum_{t=1}^n \binom {t+1}3+\binom {t+2}3\\
&=\binom {n+2}4+\binom {n+3}4\\
&=\frac 14\big[(n-1)+(n+3)\big]\binom {n+2}3\\
&=\frac 12(n+1)\binom {n+2}3\\
&=\frac {(n+2)(n+1)^2n}{12}\\
\lim_{n\to \infty}\frac 1{n^4}\sum_{r=1}^n\left[(2r-1)\sum_{k=1}^{n-r+1}k\right]
&=\lim_{n\to\infty}\frac 1{n^4}\cdot\frac {(n+2)(n+1)^2n}{12}\\
&=\lim_{n\to\infty}\frac 1{n^4}\cdot \frac {n^4+O(n^3)}{12}\\
&=\lim_{n\to\infty}\frac {1+O(\frac 1n)}{12}\\
&=\frac 1{12}
\end{align}$$
