Limit of cubes and natural numbers

$\begin{aligned} L & = \lim_{n \to \infty} \frac {n^2(1+2+3+\cdots+n)}{2(1^3+2^3+3^3+\cdots +n^3)} \\ & = \lim_{n \to \infty} \frac {n^2 \sum_{k=1}^n k}{2\sum_{k=1}^n k^3} \\ & = \lim_{n \to \infty} \frac {n^2\cdot \frac {n(n+1)}2}{2\left(\frac {n(n+1)}2\right)^2} \\ & = \lim_{n \to \infty} \frac {n^2}{n(n+1)} \\ & = \lim_{n \to \infty} \frac {n^2}{n^2+n} & \small \color{#3D99F6} \text{Divide up and down by }n^2 \\ & = \lim_{n \to \infty} \frac 1{1+\frac 1n} \\ & = \boxed{1} \end{aligned}$

The fraction may be rewritten as: n^2[(n/2)(n + 1)]/2[((n^2)/4)(n + 1)^2] =(n^2 + n)/(n^2 + 2n + 1). Dividing numerator and denominator by n^2, (1 + 1/n)/(1 + 2/n + 1/n^2). Lim as n app inf. = 1.