Question

Write the value of $\underset{n\to \infty }{\lim }\frac{1+2+3+...+n}{n^2}.$

Answer 1

$$\begin{gathered} \underset{n\to \infty }{\lim }\left(\frac{1+2+3...n}{n^2}\right) \\ =\underset{n\to \infty }{\lim }\left(\frac{n\left(n+1\right)}{2n^2}\right) \\ =\underset{n\to \infty }{\lim }\left(\frac{\left(n+1\right)}{2n}\right) \\ =\frac{1}{2}\underset{n\to \infty }{\lim }\left(1+\frac{1}{n}\right) \\ =\frac{1}{2} \end{gathered}$$