## Calculus: Early Transcendentals (2nd Edition)

The solution is $$\lim_{n\to\infty}\frac{1+2+\cdots+n}{n^2}=\frac{1}{2}.$$
To solve this limit we will use the hint given in the problem $1+2+3+\cdots+n=\frac{n(n+1)}{2}$: $$\lim_{n\to\infty}\frac{1+2+\cdots+n}{n^2}=\lim_{n\to\infty}\frac{\frac{n(n+1)}{2}}{n^2}=\lim_{n\to\infty}\frac{n^2\left(1+\frac{1}{n}\right)}{2n^2}=\lim_{n\to\infty}\frac{1}{2}\left(1+\frac{1}{n}\right)=\left[\frac{1}{2}\left(1+\frac{1}{\infty}\right)\right]=\left[\frac{1}{2}\left(1+0\right)\right]=\frac{1}{2}.$$