Answer to Question #151779 in Calculus for Cypress

Question

Determine the limit of (1+2+...+n)/(n^2) when n goes to infinity.
Can I solve this by taking the sequence to n brackets?

Answer 1

Let $a_n=\dfrac{1+2+...+n}{n^2}.$ Then

$$a_n=\dfrac{1+2+...+n}{n^2}$$

$$a_n=\dfrac{\displaystyle\bigg(\sum_{i=1}^i\bigg)}{n^2}$$

$$a_n=\dfrac{\displaystyle\dfrac{n(n+1)}{2}}{n^2}$$

$$a_n=\dfrac{n^2+n}{2n^2}$$

$$a_n=\dfrac{1}{2}+\dfrac{1}{2n}$$

$$\lim\limits_{n\to \infin}\dfrac{1+2+...+n}{n^2}=\lim\limits_{n\to \infin}a_n$$

$$=\lim\limits_{n\to \infin}\big(\dfrac{1}{2}+\dfrac{1}{2n}\big)=\dfrac{1}{2}+0$$

$$=\dfrac{1}{2}$$

Therefore

$$\lim\limits_{n\to \infin}\dfrac{1+2+...+n}{n^2}=\dfrac{1}{2}$$