Let $s > 0$ be the reserve of a company, $\pi$ be the turnover in year $t$ (that is identical in every single year) and $S_t$ be the expenses, with $S_t$ being i.i.d. Show that:
$$E(S_t) > \pi \Rightarrow P(s + T\pi - \sum_{t=1}^T S_t < 0 \text{ for some} \ T \in \Bbb N) = 1.$$
I tried the following:
$$P(s + T\pi - \sum_{t=1}^T S_t < 0 \text{ for some} \ T \in \Bbb N) = 1$$
is equivalent to
$$P(s + T\pi - \sum_{t=1}^T S_t \ge 0 \text{ for every } T \in \Bbb N) = 0,$$
which I want to show instead.
Now,
$$P(s + T\pi - \sum_{t=1}^T S_t \ge 0 \text{ for every } T \in \Bbb N) = P(s + T\pi \ge \sum_{t=1}^T S_t \text{ for every } T \in \Bbb N).$$
By dividing both sides of the inequality by $T$, the above probability is identical to:
$$P({s \over T} + \pi \ge {1 \over T} \sum_{t=1}^T S_t \text{ for every } T \in \Bbb N).$$
Now comes the critical part: I took the $\lim_{T \rightarrow \infty}$ on both sides of the inequality. I think that it might be possible to do so because on the one hand, the above inequality is supposed to hold for every $T \in \Bbb N$, and on the other hand, since it's not a strict inequality, both sides are allowed to behave equivalently in an asymptotical kind of sense. The probability above is now $\le$ the probability:
$$P(\lim_{T \rightarrow \infty}{s \over T} + \pi \ge \lim_{T \rightarrow \infty}{1 \over T} \sum_{t=1}^T S_t) = P(\pi \ge E(S_t)).$$
The right hand side of the inequality follows from the weak law of large numbers and the fact that the $S_t$ are i.i.d. Now, since $\pi < E(S_t)$ by premise, it follows that:
$$P(\pi \ge E(S_t)) = 0,$$
hence, the statement follows.
With law of large numbers we know $\pi < \Bbb{E}[S_1] = \lim_{T\rightarrow \infty}\frac{1}{T}\sum_{t=1}^T S_t$. Fix $\varepsilon \in (0, \Bbb{E}[S_1] - \pi)$ and let $(\Omega, \mathcal{A}, \Bbb{P})$ the probability space. Then we have for almost all $\omega \in \Omega$ exists a $T_{\omega}\in \Bbb{N}$:
$$\frac{1}{T_{\omega}}\sum_{t=1}^{T_{\omega}} S_t(\omega) > \pi + \varepsilon$$ (The notation $T_{\omega}$ denotes that this $T_{\omega}$ depends on $\omega$). Without loss of generality we could assume that $T_{\omega}$ is sufficiently large, such that $\varepsilon > \frac{s}{T_{\omega}}$. So
$$\frac{1}{T_{\omega}}\sum_{t=1}^{T_{\omega}} S_t(\omega) > \pi + \varepsilon > \pi + \frac{s}{T_{\omega}} $$
It follows that $$0 > s +T_{\omega} \pi - \sum_{t=1}^{T_{\omega}} S_t(\omega)$$
So for almost all $\omega$ there exists such an $T$. The claim follows.