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Find the value of$$L=\lim_{n\to\infty}\frac1{n^4}\cdot\left[1\cdot\sum_{k=1}^nk+2\cdot\sum_{k=1}^{n-1}k+3\cdot\sum_{k=1}^{n-2}k+\cdots+n\cdot1\right].$$

My attempt is as follows: $$S=\sum_{i=1}^{n}i\sum_{j=1}^{n-i+1}j$$ $$S=\sum_{i=1}^{n}i\cdot\dfrac{(n-i+1)(n-i+2)}{2}$$ $$S=\dfrac{1}{2}\sum_{i=1}^{n}i\cdot(n^2-ni+2n-in+i^2-2i+n-i+2)$$ $$S=\dfrac{1}{2}\sum_{i=1}^{n}i\cdot(n^2+i^2-2ni-3i+3n+2)$$ $$S=\dfrac{1}{2}\left(\dfrac{n^3(n+1)}{2}+\dfrac{n^2(n+1)^2}{4}-\dfrac{n(n+1)(2n+1)(2n+3)}{6}+\dfrac{n(n+1)(3n+2)}{2}\right)$$ $$S=\dfrac{n(n+1)}{4}\left(n^2+\dfrac{n(n+1)}{2}-\dfrac{(2n+1)(2n+3)}{3}+3n+2\right)$$ $$S=\dfrac{n(n+1)}{4}\left(\dfrac{6n^2+3n(n+1)-2(2n+1)(2n+3)+6(3n+2)}{6}\right)$$ $$S=\dfrac{n(n+1)}{4}\left(\dfrac{9n^2+3n-2(4n^2+8n+3)+18n+12}{6}\right)$$ $$S=\dfrac{n(n+1)}{4}\left(\dfrac{n^2+5n+6}{6}\right)$$ $$S=\dfrac{n(n+1)(n+2)(n+3)}{24}$$ $$L=\lim_{n\to\infty}\frac1{24}{\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{2}{n}\right)\left(1+\dfrac{3}{n}\right)}$$ hence answer is $\dfrac{1}{24}$.

But is there any shorter way to do this as in the last we got $S=\dfrac1{24}n(n+1)(n+2)(n+3)$, which is a nice closed expression.

Ѕᴀᴀᴅ
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user3290550
  • 3,452

2 Answers2

1

Let $$S_n=1\cdot\sum_{k=1}^nk+2\cdot\sum_{k=1}^{n-1}k+3\cdot\sum_{k=1}^{n-2}k+\cdots+n\cdot1.$$ $$S_1=1=\frac{1\cdot2\cdot3\cdot4}{24}.$$ Now, by the assumption of the induction we obtain: $$S_{n+1}=S_{n}+(n+1)+2\cdot n+3\cdot(n-1)+...+(n-1)\cdot3+n\cdot2+(n+1)\cdot1=$$ $$=\frac{n(n+1)(n+2)(n+3)}{24}+\sum_{k=1}^{n+1}(n+2-k)k=$$ $$=\frac{n(n+1)(n+2)(n+3)}{24}+(n+2)\cdot\frac{(n+1)(n+2)}{2}-\frac{(n+1)(n+2)(2n+3)}{6}=$$ $$=\frac{(n+1)(n+2)}{24}(n(n+3)+12(n+2)-4(2n+3))=$$ $$=\frac{(n+1)(n+2)(n+3)(n+4)}{24}.$$

Michael Rozenberg
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WE can write it as $$L=\lim_{n \to \infty} \frac{1}{n^4}\sum_{j=1}^{n} j \sum_{k=1}^{n-j+1} k =\lim_{n \to \infty} \frac{1}{n^4}\sum_{j=1}^{n} \frac{j(n-j+1)(n-j+2)}{2}$$ $$L = \lim_{n \to \infty} \frac{1}{2n^4}\sum_{j=1}^n(j^3-2nj^2+n^2j+....)$$ Let us use the asymptotic result when $n$ is very large: $\sum_{k=1}^n k^z \sim\frac{n^{z+1}}{z+1}$ we get $$L=\lim_{n \to \infty} \frac{1}{2n^4} (\frac{n^4}{4}-2\frac{n^4}{3}+\frac{n^4}{2})=\frac{1}{24}$$

Z Ahmed
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