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

Question

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?

Answer 1

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.