Show $\frac{1}{n^2}\sum^n_{k=1} \lfloor kx\rfloor$ where $x \in \mathbb R$ converges and calculate its limit

The question is as follows, Show that the sequence whose general term $u_n$ is given converges and calculate its limit,

$\frac{1}{n^2}\sum^n_{k=1} \lfloor kx\rfloor$ where $x \in \mathbb R$.

So the $c_n = \frac{1}{n^2}\lfloor kx\rfloor$ right?

and I have to prove that for all $\epsilon > 0$ there exists an $N$ such that if $n > N$, then $|\frac{1}{n^2}\lfloor kx\rfloor - l| < \epsilon$ where $l$ is the limit?

usually I'm given the l value and theres no floor function so...what should my next step be? provided I'm even headed in the right direction

• No, $c_n=\frac1{n^2}\sum_{k=1}^n\lfloor kx\rfloor$. – Brian M. Scott Feb 22 '13 at 11:25

Use the fact that $kx-1<\lfloor kx\rfloor\le kx$

So we have,

$$\sum_{k=1}^{n}(kx-1)<\sum_{k=1}^{n}\lfloor kx\rfloor\le \sum_{k=1}^{n}kx$$

So we have,

$$\frac{n(n+1)x}{2}-n<\sum_{k=1}^{n}\lfloor kx\rfloor\le \frac{n(n+1)x}{2}$$

$$\frac{1}{n^2}\left(\frac{n(n+1)x}{2}-n\right)<\frac{1}{n^2}\sum_{k=1}^{n}\lfloor kx\rfloor\le \frac{1}{n^2}\frac{n(n+1)x}{2}$$

$$\Rightarrow \left(\frac{1}{2}+\frac{1}{2n}\right)x-\frac{1}{n}<\frac{1}{n^2}\sum_{k=1}^{n}\lfloor kx\rfloor\le \left(\frac{1}{2}+\frac{1}{2n}\right)x$$

Take $l=x/2$ and you are done.

• Okay thank you all for your responses. Basically I have to use sandwich theorem yes? Prove that limit exists for kx-1 case and kx case and hence exists for floor kx case. – gvatar Feb 22 '13 at 11:39
• Yes you are right @gvatar you have to you use sandwitch theorem – Abhra Abir Kundu Feb 22 '13 at 11:41

Hint: $kx-1 < \lfloor kx \rfloor \le kx$, so

\begin{align} \sum_{k=1}^n (kx-1) &< \sum_{k=1}^n \lfloor kx \rfloor \le \sum_{k=1}^n kx \\ x \frac{n(n+1)}{2} - n &< \sum_{k=1}^n \lfloor kx \rfloor \le x \frac{n(n+1)}{2}. \end{align}