\displaystyle $\begin{align*}
\sum_{n = 1}^{\infty} \mathbb{P}(X^2 > n)
& = \sum_{n = 1}^{\infty} \mathbb{P}(X > \sqrt{n})
= \sum_{n = 1}^{\infty} \mathbb{P}(X \geq \lfloor \sqrt{n} \rfloor + 1)
= \sum_{n = 1}^{\infty} \sum_{k = \lfloor \sqrt{n} \rfloor + 1}^{\infty} \mathbb{P}(X = k) \\
& = \sum_{n = 1}^{\infty} \sum_{k = \lfloor \sqrt{n} \rfloor + 1}^{\infty} p (1 - p)^{k - 1}
= p \sum_{n = 1}^{\infty} \sum_{k = 0}^{\infty} (1 - p)^{k + \lfloor \sqrt{n} \rfloor + 1 - 1}
= p \sum_{n = 1}^{\infty} (1 - p)^{\lfloor \sqrt{n} \rfloor} \sum_{k = 0}^{\infty} (1 - p)^k \\
& = p \sum_{n = 1}^{\infty} (1 - p)^{\lfloor \sqrt{n} \rfloor} \frac{1}{p}
\stackrel{(*)}{=} \sum_{n = 1}^{\infty} (2 n + 1) (1 - p)^n
= ({\sf redovi})
= \frac{2 - p - p^2}{p^2}
\end{align*}