I'm looking on feedback on the following proof. Most importantly, did I make any fundamental errors in my reasoning? If not, how could I make the proof more professional, adding rigour and better styllistic choices? Is my proof interesting and creative at all in your opinion?
Idea of proof and interpretation of Borel Cantelloni:
Let $E_1, E_2,...$ be a sequence of events, then the Borel Cantelloni states that if $\sum_{n=1}^{\infty} P(E_n) < \infty $, then $P(\limsup_{n\rightarrow\infty} E_n ) = 0$, meaning that (almost surely) only finitely events can happen simultaniously. Now for some subset $S = s_1, s_2, ...$ of the integers, imagine for every $s\in S $ a corresponding (independent) coinflip that has a probability of $\frac{1}{s}$ to land on Heads. If the the reciprocals of $S$ converge, i.e. : $\sum_{n=1}^{\infty} \frac{1}{s_n} < \infty$ then almost surely, only finitely coinflips resulted in Heads. We will show that assuming that the sum of prime reciprocals converges, thus assuming only finitely coinflips result in Heads, will lead to a contradiction. We will construct a random integer $r$ in the following way: We start by setting $r = 1$. Then for each prime $p_n$, we flip the corresponding coin.For each Heads, we multiply $r$ by $p_n$, and flip that coin again. We keep flipping each coin until it land on Tails, and multiplying $r$ with $p_n$ every time we flipped a Head.
The Proof:
Let $P = p_1, p_2,...$ be the set of all primes and assume that $\sum_{n=1}^{\infty} \frac{1}{p_n} < \infty $. Let $g_n$ be a sequence of independent random geometric variables with $P(g_n = k ) = \frac{1}{p_n}^k\cdot(1-\frac{1}{p_n})$. And let $r = \prod_{n=1}^{\infty}p_n^{g_n}$. And let $E_n$ be the event that $g_n > 0$. Then by Borel-Cantelloni, we have that $g_n>0$ for only finitely many $n\in N$ a.s. Since each $g_n$ is also finite a.s., we have that $r<\infty$ a.s., thus $\sum_{n=1}^{\infty} P(r=n) = 1$ a.s. Furthermore, $P(r=1)$ corresponds to the chance that every $g_n = 0 $, thus throwing no Heads at all. And $P(r=2)$ equals the probability of throwing first a Head with the coin corresponding to $2$, and then throwing Tails for every subsequent coinflip, that is $P(r=2) = \frac{1}{2} P(r=1)$. We can apply this logic for every integer, leading to the general result $P(r=n) = \frac{1}{n} P(r=1)$ and thus: $$1 = \sum_{n=1}^{\infty} P(r=n) = \sum_{n=1}^{\infty}\frac{1}{n} P(r=1)= (\sum_{n=1}^{\infty}\frac{1}{n}) P(r=1) $$.
But this is impossible, since $ \sum_{n=1}^{\infty}\frac{1}{n} = \infty$. Thus the right-hand side of the above equation is infinite if $P(r=1)>0$ and otherwise equals $0$, but can never equal $1$. Which proves by contradiction that the sum of prime reciprocals must diverge.