Sequence of random variable does not converge almost surely (but in P ?)

Let $$(X_n)_{n \geq 1}$$ be a sequence of iid random variables such that $$\mathbb P(X_n = 0) = \frac{1}{2}$$ and $$\mathbb P(X_n = 1) = \frac{1}{2}$$.

For all $$n \geq 1$$, let $$S_n := \sum_{k=1}^n 2^{k-n-1}X_k$$ and $$T_n := \sum_{k=1}^n 2^{-k}X_k$$.

I have a question: Why is there no random variable $$S$$ such that $$S_n \to S$$ almost sure?

I already showed that $$S_{n+1} = \frac{1}{2} (X_{n+1} + S_n)$$. Now my plan is to continue by contradiction that $$S_n \to S$$ almost sure for some r.v. $$S$$, but I'm struggling a bit. Could anybody help me out? (And tell me if the solution is correct, until now?)

And why does $$S_n \to S$$ in probability?

Suppose $$S_n \to S$$ a.s. Then $$X_{n+1}=2S_{n+1}-S_n \to 2S-S=S$$ a.s. But an i.i..d sequence can converge almost surely only when the r.v.'s are a.s constant but that is not the case here.
Here is an elementary argument: Suppose $$X_n \to S$$ a.s. Then $$X_{n+1}-X_n \to 0$$ a.s and this implies $$P(|X_{n+1}-X_n| >\frac 1 2) \to 0$$. But $$P(|X_{n+1}-X_n| >\frac 1 2)\geq P(X_{n+1}=1,X_n=0)=\frac 1 4$$ by independence.
• Thanks a lot for your answer. So am I right that $S_n \to S$ not almost sure, but in probability? May 21, 2021 at 9:29
• Can you prove that $S_n$ converges in probability? @Stanisla May 21, 2021 at 9:30
• I tried to, but I'm not sure... I would say the following: Let's suppose $S_n \to S$ in probability. Then $\mathbb P(|S_n - S| \leq \epsilon)$, we can use $\epsilon = \frac{1}{2}$ and this converges to 1. But this is exaxtly the definition of "convergence in probability". Any correction / comment is appreciated... May 21, 2021 at 9:37