My Rearranging Limit

Let us denote the numerator by $\color{#D61F06}{\mathfrak{N}}$ .

$\large \displaystyle \color{#D61F06}{{\mathfrak{N}}} = \sum_{k=1}^n k^2(n-(k-1)) = n\sum_{k=1}^n k^2 - \sum_{k=1}^n k^3+\sum_{k=1}^n k^2$

$\large\color{#D61F06}{\mathfrak{N}} = \frac{1}{6}n^2(n+1)(2n+1) - \frac{n^2(n+1)^2}{4} + \frac{1}{6}n(n+1)(2n+1)$

$\large \mathfrak{T}=\displaystyle\lim_{n\to\infty}\dfrac{1^2n+2^2(n-1)+3^2(n-2)+\cdots+n^2\cdot 1}{1^3+2^3+3^3+\cdots +n^3}=\dfrac{\color{#D61F06}{\alpha}}{\color{#3D99F6}{\varphi}}$

$\displaystyle\large\implies\mathfrak{T} = \lim_{n\to\infty}\frac{\large \color{#D61F06}{{\mathfrak{N}}}}{\frac{n^2(n+1)^2}{4}}$

We divide throughout by $n^4$

$\displaystyle\large\implies \mathfrak{T} = 4\lim_{n\to\infty} \frac{\frac{1}{6}\frac{n^2(n+1)(2n+1)}{n^4} - \frac{1}{4}\frac{n^2(n+1)^2}{n^4} + \frac{1}{6}\frac{n(n+1)(2n+1)}{n^4} }{\frac{n^2(n+1)^2}{n^4}}$

$\displaystyle\large \implies \mathfrak{T} = 4\lim_{n\to\infty} \frac{\frac{1}{6}\frac{\cancel{n^2}}{\cancel{n^2}}(\frac{1}{n}+1)(\frac{1}{n}+2) - \frac{\cancel{n^2} (\frac{1}{n} + 1)^2}{4\cancel{n^2}} + \frac{1}{6n}\frac{\cancel{n}}{\cancel{n}}(\frac{1}{n}+1)(\frac{1}{n}+2)}{\frac{\cancel{n^2} (\frac{1}{n} + 1)^2}{\cancel{n^2}} }$

Applying Limit The last expression in the numerator turns zero as the power of $n$ is less than 4. , so

$\displaystyle\large \mathfrak{T} = 4 \frac{\frac{2}{6}-\frac{1}{4}}{1^2}=\frac{\color{#D61F06}{\overbrace{1}}}{\color{#3D99F6}{\underbrace{3}}}=\boxed{0.333}$

The limit can be expressed as

$\lim_{n \to \infty} \dfrac{\sum_{k=1}^{n} (k^2(n-k+1))}{\sum_{k=1}^{n} k^3}$

So we have,

$\begin{aligned} \lim_{n \to \infty} \dfrac{\sum_{k=1}^{n} (k^2(n-k+1))}{\sum_{k=1}^{n} k^3} &=& \lim_{n \to \infty} \dfrac{n\sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k^3 + \sum_{k=1}^{n} k^2}{\sum_{k=1}^{n} k^3}\\ &=& \lim_{n \to \infty} \dfrac{n\sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k^3 + \sum_{k=1}^{n} k^2}{\sum_{k=1}^{n} k^3} \div \dfrac{n^4}{n^4}\\ &=& \lim_{n \to \infty} \dfrac{\dfrac{1}{n}\sum_{k=1}^{n} \left(\dfrac{k}{n} \right)^2 - \dfrac{1}{n}\sum_{k=1}^{n} \left(\dfrac{k}{n} \right)^3 + \dfrac{1}{n^2}\sum_{k=1}^{n} \left( \dfrac{k}{n}\right)^2}{\dfrac{1}{n}\sum_{k=1}^{n} \left(\dfrac{k}{n} \right)^3}\\ &=& \dfrac{\int_{0}^{1} x^2 dx - \int_{0}^{1} x^3 dx + 0}{\int_{0}^{1} x^3 dx}\\ &=& \boxed{\dfrac{1}{3}} \end{aligned}$

Relevant wiki: Stolz–Cesàro theorem

This problem can be solved using Stolz–cesàro theorem . Let $a_n$ denote the numerator of the fraction given in the limit, similarly let $b_n$ denote the numerator of the fraction given in the limit.

By Stolz-Cesaro theorem, if the following limit exists, then it is equal to the limit in question:

$\lim_{n\to\infty} \dfrac{a_{n+1} - a_n}{b_{n+1} - b_n} \; .$

In other words, if $\displaystyle \lim_{n\to\infty} \dfrac{a_{n+1} - a_n}{b_{n+1} - b_n} = \mathcal L$ for some finite $\mathcal L$ , then $\displaystyle \lim_{n\to\infty} \dfrac{a_n}{ b_n} = \mathcal L$ .

We know that $\displaystyle b_n = \sum_{k=1}^n k ^3\Rightarrow b_{n+1} - b_n = (k+1)^3$ , and

$\begin{aligned} && a_n = 1^2(n)+2^2(n-1)+3^2(n-2)+\cdots+n^2(1) \\ && \Leftrightarrow a_{n+1} = 1^2(n+1) + 2^2(n) + 3^2(n-1) + n^2(2) + (n+1)^2 (1) \; . \end{aligned}$

So $a_{n+1} - a_n = 1^2 + 2^2 + 3^2 + \cdots + n^2 + (n+1)^2$ , thus

$\begin{aligned} \lim_{n\to\infty} \dfrac{a_{n+1} - a_n}{b_{n+1} - b_n} &=& \lim_{n\to\infty} \dfrac{1^2+2^2+3^2+ \cdots + (n+1)^2}{(n+1)^3} \\ &= & \lim_{n\to\infty} \dfrac{1^2+2^2+3^2+ \cdots + n^2}{n^3} \\ &= & \lim_{n\to\infty} \dfrac1n\sum_{j=1}^n \left( \dfrac jn \right)^2 \\ &= & \int_0^1 x^2 \, dx = \dfrac13 \; . \end{aligned}$

Since the limit exists, then so is the limit in question. Hence our answer is $\boxed{\dfrac13}$ .

$\begin{aligned} L & = \lim_{n \to \infty} \frac {1^2(n)+2^2(n-1)+3^2(n-2)+...+n2(1)}{1^3+2^3+3^3+...+n^3} \\ & = \lim_{n \to \infty} \frac {\sum_{k=1}^n k^2(n+1-k)}{\sum_{k=1}^n k^3} \\ & = \lim_{n \to \infty} \frac {(n+1)\sum_{k=1}^n k^2 - \sum_{k=1}^n k^3 )}{\sum_{k=1}^n k^3} \\ & = \lim_{n \to \infty} \frac {(n+1)\sum_{k=1}^n k^2}{\sum_{k=1}^n k^3} - 1 \\ & = \lim_{n \to \infty} \frac {(n+1)\cdot \frac {n(n+1)(2n+1)}6}{\frac {n^2(n+1)^2}4} - 1 \\ & = \lim_{n \to \infty} \frac {4(2n+1)}{6n} - 1 \\ & = \lim_{n \to \infty} \frac {2(2+\frac 1n)}3 - 1 \\ & = \frac 43 - 1 = \frac 13 \approx \boxed{0.333} \end{aligned}$