Let be consider random variables $X_i$ which are independent and such that $\Bbb{P}(X_i=1)=1-2^{-i}$ and $\Bbb{P}(X_i=1-2^i)=2^{-i}$. Then define $S_n=\sum_{k=1}^n X_k$ and $T=\min\{n\geq 0: S_n=1\}$.
I want to check that $T<\infty $ a.s.
I have already remarked that $S_n$ is a martingale. My idea was to show that $\Bbb{P}(T=\infty)=0$. Then $$\Bbb{P}(T=\infty)=\Bbb{P}(\min\{n\geq 0: S_n=1\}=\infty)=\Bbb{P}(S_n\neq 1~\forall n)=1-\Bbb{P}(\exists~ n~: S_n=1)$$Now I somehow wanted to rewrite the event that for one $n$, $S_n=1$. but I don't managed to do it, is there a trick?
Note $$\sum_{n \in \mathbb{N}}P(X_n \neq 1)=\sum_{n \in \mathbb{N}}2^{-n}<\infty$$ So by Borel-Cantelli I $P(X_n\neq 1\textrm{ i.o.})=0$. So in particular $$P\bigg(\bigcup_{n \in \mathbb{N}}\bigcap_{k\geq n}\{X_k=1\}\bigg)=1$$ I.e. for a.a. $\omega \in \Omega$, $\exists n(\omega)$ s.t. $\forall k\geq n(\omega)$, $X_k(\omega)=1$. So $S_n$ eventually 'shots up' to infinity, almost surely, and reaches any positive integer. So $P(T<\infty)=1$.