Solution
= Solution
Pairing the first and last terms of the <arithmetic progression> gives $\sum_{k=1}^n k=n(n+1)/2$. Therefore the <limit of a sequence> is
$$
\frac{1+2+\cdots+n}{n^2}
=\frac12+\frac1{2n}\longrightarrow\boxed{\frac12}.
$$
= Solution
Pairing the first and last terms of the <arithmetic progression> gives $\sum_{k=1}^n k=n(n+1)/2$. Therefore the <limit of a sequence> is
$$
\frac{1+2+\cdots+n}{n^2}
=\frac12+\frac1{2n}\longrightarrow\boxed{\frac12}.
$$