How to prove floor function inequality $\sum\limits_{k=1}^{n}\frac{\{kx\}}{\lfloor kx\rfloor }<\sum\limits_{k=1}^{n}\frac{1}{2k-1}$ for $x>1$ 
Let $x>1$ be a real number. Show that for any positive $n$
$$\sum_{k=1}^{n}\dfrac{\{kx\}}{\lfloor kx\rfloor }<\sum_{k=1}^{n}\dfrac{1}{2k-1}\tag{1}$$
  where $\{x\}=x-\lfloor x\rfloor$

My attempt: I try use induction prove this inequality.
It is clear for $n=1$, because $\{x\}<1\le \lfloor x\rfloor$.
Now if assume that $n$ holds, in other words: $$\sum_{k=1}^{n}\dfrac{\{kx\}}{\lfloor kx\rfloor }<\sum_{k=1}^{n}\dfrac{1}{2k-1}$$
Consider the case $n+1$. We have
$$\sum_{k=1}^{n+1}\dfrac{\{kx\}}{\lfloor kx\rfloor }=\sum_{k=1}^{n}\dfrac{\{kx\}}{\lfloor kx\rfloor }+\dfrac{\{(n+1)x\}}{\lfloor (n+1)x\rfloor}<\sum_{k=1}^{n}\dfrac{1}{2k-1}+\dfrac{\{(n+1)x\}}{\lfloor (n+1)x\rfloor}$$
It suffices to prove that
$$\dfrac{\{(n+1)x\}}{\lfloor (n+1)x\rfloor}<\dfrac{1}{2n+1}\tag{2}$$
But David gives an example showing $(2)$ is wrong, so how to prove $(1)$?

Some thoughts:
Looking at several plots indicates that $$f_n(x):=\sum_{k=1}^n{\{kx\}\over\lfloor kx\rfloor}$$
is largest immediately to the left of $x=2$. Now for $x=2-\epsilon$ with $0<\epsilon\ll1$ one has
$$\lfloor kx\rfloor=2k-1,\quad\{kx\}=1-2k\epsilon$$
and therefore
$$f_n(x)=\sum_{k=1}^n{1-2k\epsilon\over2k-1}<\sum_{k=1}^n{1\over2k-1}\ .$$
Maybe you want to take a look at the following graph of $f_{250}$:

