Prove $\Sigma_{k=0}^{n-1}\lfloor x+\frac{k}{n}\rfloor=\lfloor nx\rfloor$ , n is a Natural Number

Prove the following identity: $$\lfloor x\rfloor +\lfloor x+\frac{1}{n}\rfloor +\lfloor x+\frac{2}{n}\rfloor +\lfloor x+\frac{3}{n}\rfloor+...+\lfloor x+\frac{n-1}{n}\rfloor =\lfloor nx\rfloor$$ where n is a Natural Number.

At first I thought of splitting it into 2 cases: when x is an integer, and when x isnn't an integer. The case of $x$ being an integer is quite simple: $\lfloor x+\frac{k}{n}\rfloor=x$ for $0\le k\le n-1$. Thus the LHS becomes $n\lfloor x\rfloor$ which is equal to the RHS. However I do not know how to go about the case of $x$ not being an integer.

Lastly, I would actually prefer a proof where it is not necessary to make cases based on the values of $x$, but to have one general proof which satisfies all $x$.

Could somebody please show me how to complete the proof in both ways (ie first, the case where $x$ isn't an integer, and secondly the general proof which doesn't involve breaking into cases)? Many thanks in advance!

• $\lfloor y \rfloor = y - \{y\}$ where $\{y\}$ is the fractionnal part : $1$-periodic, and $\{y\}-1/2$ is odd and zero-mean – reuns Jun 6 '16 at 1:48
• Sir, I haven't yet done Periodic Functions although I have an idea of what they are. COuld you please elaborate your method? – user342209 Jun 6 '16 at 1:49
• We can write $x = m + \delta$ where $m$ is an integer and $\delta \in [\frac{k}{n}, \frac{k+1}{n})$ for some integer $0\leq k < n$. – Winther Jun 6 '16 at 1:49
• Sir, how does that help? Could you please post a solution? – user342209 Jun 6 '16 at 1:51
• no, this forum is not for solving your exercices. so try with the hints we gave you, or tell us where you are stuck – reuns Jun 6 '16 at 1:51

Choose integers $a,b$ such that $0<b\le n$ and $$a-\frac bn\le x<a-\frac {b-1}n\ .$$ (In other words, round $x$ downwards to the nearest multiple of $1/n$.) Then $$\Bigl\lfloor x+\frac kn\Bigr\rfloor =\cases{a-1&if k=0,1,\ldots,b-1\cr a&if k=b,b+1,\ldots,n-1.}$$ So $$LHS=b(a-1)+(n-b)a=na-b\ ;$$ on the other hand, $$na-b\le nx<na-b+1$$ so $$RHS=na-b=LHS\ .$$
Let me indicate a more continuous process for David's answer $$\begin{gathered} \sum\limits_{k = 0}^{n - 1} {\left\lfloor {x + \frac{k} {n}} \right\rfloor } = n\left\lfloor x \right\rfloor + \sum\limits_{k = 0}^{n - 1} {\left\lfloor {\left\{ x \right\} + \frac{k} {n}} \right\rfloor } = n\left\lfloor x \right\rfloor + \sum\limits_{k\, \geqslant \,\,n\,\left( {1 - \left\{ x \right\}} \right)}^{n - 1} 1 = \hfill \\ = n\left\lfloor x \right\rfloor + n - \left\lceil {n\,\left( {1 - \left\{ x \right\}} \right)} \right\rceil = n\left\lfloor x \right\rfloor - \left\lceil {\, - n\left\{ x \right\}} \right\rceil = n\left\lfloor x \right\rfloor + \left\lfloor {n\left\{ x \right\}} \right\rfloor = \hfill \\ = \left\lfloor {nx} \right\rfloor \hfill \\ \end{gathered}$$