Some sequence of natural numbers

$\begin{aligned} L & = \lim_{n \to \infty} \left(\frac 1{n^2} + \frac 2{n^2} + \frac 3{n^2} + \cdots + \frac {n-1}{n^2} \right) \\ & = \lim_{n \to \infty} \frac 1{n^2} \sum_{k=1}^{n-1} k \\ & = \lim_{n \to \infty} \frac 1{n^2} \times \frac {(n-1)n}2 \\ & = \lim_{n \to \infty} \frac {n^2-n}{2n^2} & \small \color{#3D99F6} \text{Divide up and down by }n^2 \\ & = \lim_{n \to \infty} \frac {1-\frac 1n}2 \\ & = \frac 12 = \boxed{0.5} \end{aligned}$

an alternate method is to recognize this is a Riemann sum, i.e $\lim_{n\to \infty} \dfrac{1}{n} \sum_{k=1}^{n-1} \dfrac{k}{n} = \int_0^1 x \text{dx} = \dfrac{1}{2}$

As the common denominator $n^2$ , We can add all the numbers from 1 to $n-1$ in the numerator simply arithmetic sum formula : $S = \frac{(1 + (n - 1))\cdot n }{2} )$

$S = \frac{n^2}{2}$

$lim_{x\to\infty} \frac{S}{n^2}$ = $lim_{x\to\infty} \frac{\frac{n^2}{2}}{n^2}$

Cancel out $n^2$

$lim_{x\to\infty} \frac{1}{2} = \boxed{\frac{1}{2}}$

Simply, the sum above the line of the fraction is the sum of the first n-1 natural numbers. After the reduction, dividing by n, both parts of the ratio, we come to the result of 1/2.

$\begin{aligned} &\textcolor{#FFFFFF}{=}\lim\limits_{n\rightarrow\infin}\left(\frac{1}{n^2}+\frac{2}{n^2}+\cdots+\frac{n-1}{n^2}\right)\\ &=\lim\limits_{n\rightarrow\infin}\frac{1+2+\cdots+n-1}{n^2}\\ &=\lim\limits_{n\rightarrow\infin}\frac{1}{n^2}\frac{(n-1)(n-2)}{2}\\ &=\lim\limits_{n\rightarrow\infin}\frac{n^2-3n+2}{2n^2}\\ &=\lim\limits_{n\rightarrow\infin}\frac{1-\frac3n+\frac{2}{n^2}}{2}\\ &=\frac{1-0+0}{2}\\ &=\boxed{\frac12} \end{aligned}$