Given the Lindeberg-Lévy Central Limit Theroem for i.i.d $X_i$'s,
\begin{equation} \sqrt{n}\left(\frac{1}{n}\sum_{i=1}^{n}X_i-\mu\right)\overset{d}\rightarrow N(0,\sigma^2) \end{equation}
I understand this implies
\begin{equation} \frac{1}{n}\sum_{i=1}^{n}X_i-\mu = O_p\left(\frac{1}{\sqrt{n}}\right) \end{equation}
so that when $\mu=0$, we have the following boundedness in probability \begin{equation} \frac{1}{\sqrt{n}}\sum_{i=1}^{n}X_i = O_P(1)~\Rightarrow~ \sum_{i=1}^{n}X_i = O_p(\sqrt{n}). \end{equation}
Question. What if $\mu$ is not necessarily zero? What can we say about the order of $\sum_{i=1}^{n}X_i$? Am I correct in saying that
\begin{equation} \frac{1}{n}\sum_{i=1}^{n}X_i = O_p\left(\frac{1}{\sqrt{n}}\right) + \mu = O_P(1) ~\Rightarrow~ \sum_{i=1}^{n}X_i = O_p(n) \end{equation}
Can we have a better bound than $O_p(n)$?