Would it be possible for me to write and if yes could somebody please help me out:
$$\sum_{i=1}^{\lfloor n\{x\}\rfloor}\frac{1}{\lfloor x+1 \rfloor^k} +\sum_{i=\lfloor n\{x\}\rfloor+1}^{n}\frac{1}{\lfloor x \rfloor^k}=\sum_{i=1}^{n}\frac{1}{\lfloor x+ \frac{i-1}{n}\rfloor^k}$$
My main problem is to include $i$ and $n$ in the sum. I would be really thankful for help.
P.S Please, I only want hints, not solutions. I would really like to solve it on my own, it's just that in this specific point i am stuck for days now.
Edit: $\{x\}=x-\lfloor x \rfloor$
Hint: $$1 \le i \le n \implies 0 \le \frac{i-1}{n} \le \frac{n-1}{n} < 1,$$ so $$x \le x+\frac{i-1}{n} \le x+\frac{n-1}{n} < x+1$$ and $$\lfloor x \rfloor \le \left\lfloor x+\frac{i-1}{n} \right\rfloor \le \left\lfloor x+\frac{n-1}{n} \right\rfloor \le \lfloor x+1 \rfloor,$$ hence $$\left\lfloor x+\frac{i-1}{n} \right\rfloor \in \{\lfloor x \rfloor, \lfloor x+1 \rfloor\}.$$
By the way, simplifying the sum of the floor powers (rather than their reciprocals) was the first part of Mathematics Magazine Problem 1785 (December 2007), and the same technique applies here.