Inequality involving size of random variable I’m following my professor’s notes, and I became confused by an inequality he used without justification, but I cannot see why it is true. Any help is appreciated.
Let $\{T_t\}_t$ be a sequence of iid, positive, integer-valued random variables. Suppose $$P\{T \ge x \} = \Theta \left(\frac{1}{\sqrt x}\right). $$
What I do not understand is he then states:
$$(1) \max_{1 \leq i \leq t} T_i = \Theta (t^2),$$
$$ (2) \sum_{i=1}^t T_i = \Theta (t^2), $$
both in probability. I would appreciate help on how he concluded both of these from the previous fact.
Thank you!

Say $ a/\sqrt{x} \le P(T \ge x) \le b/\sqrt{x}$ for $x$ sufficiently large.
For (1):
$$ P\left(\max_{1\le i \le t} T_i < c t^2\right) =  P(T < c t^2)^t $$
When $c > 0$, for $t$ sufficiently large this is between $(1 - a/(\sqrt{c} t))^t$
and $(1 - b/(\sqrt{c} t))^t$, which as $t \to \infty$ go to $\exp(-a/\sqrt{c})$ and
$\exp(-b/\sqrt{c})$ respectively.  Thus for any $\epsilon > 0$, taking $c_1$ and $c_2$ such that $\exp(-a/\sqrt{c_1}) < \epsilon$ and $\exp(-b/\sqrt{c_2}) > 1-\epsilon$, we find that
$P\left(\max_{1\le i \le t} T_i < c_1 t^2\right) < \epsilon$
while  $P\left(\max_{1\le i \le t} T_i < c_2 t^2\right) > 1-\epsilon$.
