To evaluate the limit

$\lim_{n \to \infty} \frac{1 + 2 + 3 + \cdots + n}{n^2},$

we can proceed as follows:

Step 1: Express the sum of the first $n$ integers

The sum of the first $n$ integers is given by the formula: $1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}.$

Step 2: Rewrite the fraction

Substitute the sum into the fraction: $\frac{1 + 2 + 3 + \cdots + n}{n^2} = \frac{\frac{n(n+1)}{2}}{n^2}.$

Simplify: $\frac{\frac{n(n+1)}{2}}{n^2} = \frac{n(n+1)}{2n^2} = \frac{n+1}{2n}.$

Step 3: Simplify further and take the limit

Divide both numerator and denominator of $\frac{n+1}{2n}$ by $n$: $\frac{n+1}{2n} = \frac{1 + \frac{1}{n}}{2}.$

As $n \to \infty$, the term $\frac{1}{n} \to 0$. Therefore: [ \lim_{n \to \in