Sum of Indicators: $\sum_{n=1}^{\infty} \mathbb{1}\{X_1^2/\alpha^2 \geq n\} = \left(1 + \left\lfloor\frac{X_1^2}{\alpha^2}\right\rfloor\right)$ I was working on proving the following proposition:

Suppose $(X_n)$ is some iid sequence of random variables such that $\mathbb{E}(X_1)^2 < \infty$. Then for every $\alpha > 0$ we have
$$\mathbb{P}\left(\limsup_{n \to \infty}|X_n| \geq \alpha\sqrt{n}\right) = 0.$$

Proof. We wan't to invoke the Borel-Cantelli Lemma (B.C.), so since $X_n$ are iid we have
$$\sum_{n=1}^{\infty}\mathbb{P}(|X_n| \geq \alpha\sqrt{n}) = \sum_{n=1}^{\infty}\mathbb{P}(X_n^2 \geq \alpha^2 n) = \sum_{n=1}^{\infty}\mathbb{P}\left(\frac{X_1^2}{\alpha^2} \geq n\right).$$
Using the Monotone Convergence Theorem, we can represent the right-most term as
$$ \sum_{n=1}^{\infty}\mathbb{P}\left(\frac{X_1^2}{\alpha^2} \geq n\right) =  \sum_{n=1}^{\infty}\mathbb{E}\left[\mathbb{1}\left\{\frac{X_1^2}{\alpha^2} \geq n\right\}\right] = \mathbb{E}\sum_{n=1}^{\infty} \mathbb{1}\left\{\frac{X_1^2}{\alpha^2} \geq n\right\}.$$
Therefore,
$$\mathbb{E}\sum_{n=1}^{\infty} \mathbb{1}\left\{\frac{X_1^2}{\alpha^2} \geq n\right\} = \mathbb{E}\left(1 + \left\lfloor\frac{X_1^2}{\alpha^2}\right\rfloor\right) \leq 1 + \frac{\mathbb{E}X_1^2}{\alpha^2} < \infty.  \tag{1}$$
So, we can invoke B.C. and are done. The bit about the indicator equality involving the floor function is given as a hint without going into detail.
Question.  How do we get the equality involving the floor in (1)? Specifically, I don't see how
$$\sum_{n=1}^{\infty} \mathbb{1}\left\{\frac{X_1^2}{\alpha^2} \geq n\right\} = 1 + \left\lfloor\frac{X_1^2}{\alpha^2}\right\rfloor.$$
Could someone expand on this derivation please?
 A: I think there is a slight mistake. (I am interpreting $\left\lfloor x \right\rfloor$ as the greatest integer less than or equal to $x$).  If the sums are  from $0$ to $\infty$ then you get the value as stated, but otherwise you will get $\left\lfloor\frac{X_1^2}{\alpha^2}\right\rfloor$ instead of $\left\lfloor\frac{X_1^2}{\alpha^2}\right\rfloor +1$
Suppose $\left\lfloor\frac{X_1^2}{\alpha^2}\right\rfloor=k$. Then $k \leq \left\lfloor\frac{X_1^2}{\alpha^2}\right\rfloor <k+1$. So  the only integers $n$ such that $\frac{X_1^2}{\alpha^2}\geq n$ are $n=1,2,...,k$. So when you add the indicators on LHS you see that $k$ terms are $1$ and the rest are zeros. Hence the sum is exactly $k$.
A: I guess there's a small mistake in the end. Fix any $\omega \in \Omega$ . Then, $\frac{X_1(\omega)^2}{\alpha^2}$ is a finite quantity, so $\exists~ N_\omega \in \mathbb{N}$ such that $$\frac{X_1(\omega)^2}{\alpha^2} \geqslant  N_\omega\quad~\text{ and }~\quad \frac{X_1(\omega)^2}{\alpha^2} < N_\omega +1$$ and hence, $\left\lfloor\frac{X_1(\omega)^2}{\alpha^2}\right\rfloor = N_\omega$
Thus, for $n = 1 , 2 , \cdots , N_\omega$ , we have : $\mathbb{1}\left\{\frac{X_1(\omega)^2}{\alpha^2} \geq n\right\} = 1$ and otherwise, $0$ .
Therefore, $$\sum_{n=1}^{\infty}\mathbb{1}\left\{\frac{X_1(\omega)^2}{\alpha^2} \geq n\right\} = \sum_{n=1}^{N_\omega}\mathbb{1}\left\{\frac{X_1(\omega)^2}{\alpha^2} \geq n\right\} = N_\omega = \left\lfloor\frac{X_1(\omega)^2}{\alpha^2}\right\rfloor$$
Since this result is true for any $\omega \in\Omega$ , therefore : $$\sum_{n=1}^{\infty} \mathbb{1}\left\{\frac{X_1^2}{\alpha^2} \geq n\right\} = \left\lfloor\frac{X_1^2}{\alpha^2}\right\rfloor$$

It is to be noted that even after this correction, Borel - Cantelli lemma works fine, so you'll obtain what you wanted. Hope it helps.
