Calculating $\lim\limits_{n\to\infty}\frac{1+2+3+...+n-1}{n^2}$

Question

My first attempt was using limit arithmetic, but it fails because one of the operands is infinite, so that didn't work.

I then tried using the squeeze theorem:

$$b_n = \frac{1+2+3+...+n-1}{n^2}$$ $$a_n = \frac{1\cdot(n-1)}{n^2}$$ $$c_n = \frac{(n-1)\cdot(n-1)}{n^2}$$ $$\Downarrow$$ $$a_n < b_n < c_n$$

Problem is, the limit of $a_n$ is zero, while the limit of $c_n$ is one — while this narrows it down, it doesn't quite help me get the limit of $b_n$. How can I make it work?

Answer 1

Since $1+2+3+\cdots+(n-1)=\dfrac{(n-1)n}2=\dfrac{n^2-n}2$, you have:$$\lim_{n\to\infty}\frac{1+2+3+\cdots+(n-1)}{n^2}=\lim_{n\to\infty}\frac{n^2-n}{2n^2}=\frac12.$$

Answer 2

Just visualization of approach:

Answer 3

Observe that $$\begin{align*} \lim\limits_{n \rightarrow \infty} \frac {1+2+ \cdots + (n-1)} {n^2} & = \lim\limits_{n \rightarrow \infty} \frac {1} {n} \sum\limits_{r=0}^{n-1} \left ( \frac {r} {n} \right ). \\ & = \int_{0}^{1}x\ dx. \\ & = \frac {1} {2}. \end{align*}$$

Answer 4

Hint:

$$\frac{0+1+2+3+...+n-1}{n}$$ is the average value of the $n$ first positive integers. What could it be ?