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.

$$$$At first I thought of splitting it into 2 cases: when x is an integer, and when x isnn't an integer. The case of $x$ being an integer is quite simple: $\lfloor x+\frac{k}{n}\rfloor=x$ for $0\le k\le n-1$. Thus the LHS becomes $n\lfloor x\rfloor$ which is equal to the RHS.
$$$$However I do not know how to go about the case of $x$ not being an integer.
Lastly, I would actually prefer a proof where it is not necessary to make cases based on the values of $x$, but to have one general proof which satisfies all $x$.
Could somebody please show me how to complete the proof in both ways (ie first, the case where $x$ isn't an integer, and secondly the general proof which doesn't involve breaking into cases)? Many thanks in advance!
 A: Choose integers $a,b$ such that $0<b\le n$ and
$$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\ .$$
A: Let me indicate a more continuous process for David's answer
$$
\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} 
$$
