TITLE: Show that $\lim_{n\to\infty}\frac{1}{n^4}\sum_{j=1}^{n}\left((2j-1)\sum_{k=1}^{n+1-j}k\right)=\frac{1}{12}$
QUESTION [8 upvotes]: 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.

REPLY [2 votes]: 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