Basic inequality on a floor function (Proof verification). Let $a_n=n/c$ be a real sequence with $n>c\geq 1$ constant. I am inspecting if the following (surprising) result holds:
$$\bigg\lvert \frac{1}{\left \lfloor{a_n}\right \rfloor }-\frac{1}{a_n}\bigg\rvert\leq \frac{C}{n^2}, \quad \text{ for some }C>0 \text{ and all } n\in\mathbb{N}.$$
Could you please check if my proof is acceptable?
My attempt
Let $n\in\mathbb{N}:n>1$. Then
\begin{align}
a_n-1<\left \lfloor{a_n}\right \rfloor\leq a_n\iff \frac{1}{a_n-1}-\frac{1}{a_n}>\frac{1}{\left \lfloor{a_n}\right \rfloor }-\frac{1}{a_n}\geq 0.
\end{align}
By definition 
$$\frac{1}{a_n-1}-\frac{1}{a_n}=\frac{c^2}{n^2-cn}.$$
But the following sequence converges,
$$\frac{c^2}{n^2-cn}n^2=\frac{c^2}{1-c/n}\to c^2 \text{ as } n\to\infty,$$
and so it is bounded by, say, $M>0$ (for all $n$). From these facts, we have
$$\frac{1}{\left \lfloor{a_n}\right \rfloor }-\frac{1}{a_n}<\frac{M}{n^2}.$$
Identifying $M=C$, we are done.
Thanks in advance!

Your statement is not quite correct. The issue is that for $n \lt c$ you get in those cases, as angryavian's question comment has indicated for an earlier version of your question, that $\frac{1}{\lfloor n/c \rfloor}$ becomes $\frac{1}{0}$. 
I believe the start of the problem with your proof is in your line of
$$a_n-1<\left \lfloor{a_n}\right \rfloor\leq a_n\iff \frac{1}{a_n-1}-\frac{1}{a_n}>\frac{1}{\left \lfloor{a_n}\right \rfloor }-\frac{1}{a_n}\geq 0 \tag{1}\label{eq1A}$$
The issue is it implicitly assumes that $a_n - 1 \gt 0 \iff a_n \gt 1$.
One way to handle this is to add to your statement of
$$\bigg\lvert \frac{1}{\left \lfloor{a_n}\right \rfloor }-\frac{1}{a_n}\bigg\rvert\leq \frac{C}{n^2}, \quad \text{ for some }C>0 \text{ and all } n\in\mathbb{N} \tag{2}\label{eq2A}$$
something like "$\text{ where } n \ge c$". Also, in your proof, you can add mention that the case where $n = c$, so $a_n = 1$, is true as the LHS of your statement becomes $0$. Then, with your proof technique, as I stated above, can handle all of the remaining cases where $a_n \gt 1$.
