The first thing to note is that the summation can be written as
$$\frac{1}{n^3(n+1)} \sum_{i=1}^{n}i^3$$
which greatly simplifies the problem to evaluating the function
$$f(k,n) = \sum_{i = 1}^{n} i^k$$
at $k = 3$, i.e. evaluate $f(3,n)$. Luckily for us there is a nice recurrence relation that solves the problem, notice that
$$(n+1)^{k} - n^k = \sum_{i=0}^{k-1}\binom{k}{i}n^i$$
implies that
$$\frac{\Delta f}{\Delta y}\bigg|_{(k,m)} =f(k,m+1)-f(k,m)=\sum_{n=0}^{n=m}(n+1)^{k} - n^{k}= \sum_{n=0}^{n=m}\sum_{i=0}^{k-1}\binom{k}{i}n^i = \sum_{i=0}^{k-1}\binom{k}{i} f(i,m)$$
which implies that
$$\sum_{j=1}^{j=m-1}\frac{\Delta f}{\Delta y}\bigg|_{(k,j)} = f(k,m) = \sum_{j=1}^{j=m-1} \sum_{i=0}^{k-1}\binom{k}{i} f(i,j)$$
We now solve recursively, we know that $f(0,n) = \sum_{i=1}^n i^0= n $ which allows us to solve for $k=1$
$$f(1,n) = \sum_{j=1}^{j=m-1} \sum_{i=0}^{0}\binom{1}{i} f(i,j) = \sum_{j=1}^{j=m-1} f(0,j) = \sum_{j=1}^{j=m-1} j = \binom{m}{2}$$
Solving for $k=2$ we get
$$ f(2,m) = \sum_{j=1}^{j=m-1} \sum_{i=0}^{2-1}\binom{2}{i} f(i,j) = \sum_{j=1}^{j=m-1} \left( \binom{2}{0} j \binom{2}{1} \binom{j}{2} \right) = \frac{n(n+1)(2n+1)}{6}$$
performing a similar computation for $k=3$ we get
$$f(3,n) = \left(\frac{n(n+1)}{2}\right)^2$$
and therefore the limit is
$$\lim_{n \to \infty} A_n = \lim_{n \to \infty} \frac{1}{n^3(n+1)} \left(\frac{n(n+1)}{2}\right)^2\lim_{n \to \infty} \frac{n+1}{4n} = \frac{1}{4}$$
