TITLE: How to show $\lim_{n \to \infty} a_n = \frac{ [x] + [2x] + [3x] + \dotsb + [nx] }{n^2} = x/2$?
QUESTION [21 upvotes]: This question came from the prelim exam I took last month. I have a proof that seems a bit unwieldy to me (posted as an answer), so I'm opening it up to ask if there are other ways of showing this.

Let $x$ be any positive real number, and define a sequence $\{a_n\}$ by
  $$ a_n = \frac{ [x] + [2x] + [3x] + \dotsb + [nx] }{n^2} $$
  where $[x]$ is the largest integer less than or equal to $x$. Prove that $\displaystyle{\lim_{n \to \infty} a_n = x/2}$.

REPLY [4 votes]: By definition of the floor function,
$$kx-1\le\lfloor kx\rfloor\le kx.$$
Then summing on $k$ from $1$ to $n$,
$$\frac{n(n+1)}2x-n\le\sum_{k=1}^n \lfloor kx\rfloor\le \frac{(n+1)n}2x.$$
Dividing by $n^2$ and taking the limit,
$$\frac x2\le\frac1{n^2}\sum_{k=1}^\infty \lfloor kx\rfloor\le \frac x2.$$