I'm studying a paper where the authors stated preliminarily:
$$\mid R_{i,T}\mid \leq \frac{C \theta_0}{K(i_\theta,\theta,T)}, \forall i\in\{1,\dotsc,\theta^x\}$$
where $C$ is a finite contant; $\theta_0$ a positive integer; $R_{i,T}:=R(i,T)$; and $i_\theta=i-\theta\left \lfloor{\frac{i-1}{\theta}}\right \rfloor $ ; and $\theta^x$ is the least common multiple of $\theta$ and $\theta_0$; and $\theta \in \{1,\dotsc,\Theta_T\}$ integer. Note that $i_\theta \in \{1,\dotsc,\theta\}$. However, it was given that $K(i_\theta,\theta,T)=\left \lfloor{\frac{T-i_\theta}{\theta}}\right \rfloor +1$, which implies that this function can only be either $\left \lfloor{\frac{T}{\theta}}\right \rfloor$ or $\left \lfloor{\frac{T}{\theta}}\right \rfloor +1$. It also implies that $K(i_\theta,\theta,T)=O\big(\frac{T}{\theta}\big)$. Untill here it was everything fine to me. But, from this, they claimed that:
$$\mid R_{i,T}\mid \leq \frac{C\Theta_T}{T}, \forall i\in\{1,\dotsc,\theta^x\}$$
Is it right?
It is true that $\frac{C \theta_0}{K(i_\theta,\theta,T)}\leq \frac{C}{\left \lfloor{\frac{T}{\theta}}\right \rfloor}$. On one hand, $\frac{C}{\left \lfloor{\frac{T}{\theta}}\right \rfloor} \geq \frac{C}{\frac{T}{\theta}}$ (and then cannot be used), on the other, $\frac{C}{\frac{T}{\theta}} \leq\frac{C\Theta_T}{T}$. How they simply dropped the floor function and stablished the that upper bound? Even asymptotically, I don't see that $\frac{C}{\left \lfloor{\frac{T}{\theta}}\right \rfloor} = \frac{C}{\frac{T}{\theta}}$ does hold.
The asymptotics here respects $T \rightarrow \infty$, $\Theta_T \rightarrow \infty$ at a slower rate than $T$ (implying $\mid R_{i,T}\mid \rightarrow 0$).
Reference: Proof of 'lemma a.4'
Thanks in advance!