$$\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}$$​ is equivalent to

Correct option is D

​$$\begin{aligned}&\text{\textbf{Sum of First } n \text{ Natural Numbers}} \\&\quad 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2} \\[10pt]&\text{\textbf{Rewrite the Limit}} \\&\quad \lim_{n \to \infty} \frac{1 + 2 + \cdots + n}{n^2} = \lim_{n \to \infty} \frac{\frac{n(n+1)}{2}}{n^2} = \lim_{n \to \infty} \frac{n(n+1)}{2n^2} \\[10pt]&\text{\textbf{Simplify the Expression}} \\&\quad \lim_{n \to \infty} \frac{n^2\left(1 + \frac{1}{n}\right)}{2n^2} = \lim_{n \to \infty} \frac{1 + \frac{1}{n}}{2} \\[10pt]&\text{\textbf{Evaluate the Limit}} \\&\quad \text{As } n \to \infty, \frac{1}{n} \to 0 \text{, therefore:} \\&\quad \frac{1 + 0}{2} = \boxed{\frac{1}{2}}\end{aligned}$$​​