# 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.

• If summation can be made by adding some terms then it becomes easy but not able to do so Sep 19, 2017 at 4:04
• Coincidentally it is equal to the sum $-(1+2+3+4+\cdots)$ :) Sep 19, 2017 at 16:48

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.

Hint:

You need to compute $$\lim_{n\to \infty} \frac{1}{n^4} \sum_{a=0}^{n-1} \frac{(2a+1) (n-a)(n+1-a)}{2}=\lim_{n \to \infty}\frac{n (n+1)^2(n+2)}{12 n^4}.$$

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

• This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. Oct 25, 2017 at 12:01
• @MarkFrantini limit of a sum is a standard formula and it need not be stated every time it is used Oct 25, 2017 at 12:46
• @MarkFrantini I have edited it.hope it is better Oct 25, 2017 at 13:00
• Why downvote? Please explain. Oct 28, 2017 at 17:57

\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}