Suppose that $(X_n)_{n\ge1}$ is a sequence of independent random variables with $E[|X_n|] < \infty$ for all $n$ and $E(X_n) = \mu$. Prove that
$$\sum_{n=1}^{\infty}\frac{1}{2^n}X_n = \mu \; a.s$$
I am stuck with this question and not sure how to go about it. I have proven that the sum converges absolutely almost surely but am not sure if this is useful towards my goal.
Your claim does not hold without further assumptions.
To see this, first note that we can replace $X_n$ by $X_n - \mu$ and thus assume $\mu = 0$ without loss of generality (this uses $\sum_{n=1}^\infty 2^{-n}= 1$).
Let us assume that your claim is true (for any such sequence).
Now, let $f := \sum_{n=1}^\infty 2^{-n} X_n$. By assumption that your theorem is true, we see $f \equiv 0$ almost surely.
Also, set $Y_1 := 2 X_1$ and $Y_n := X_n$ for $n \geq 2$. Note that the sequence $(Y_n)_n$ also satisfies the requirements of your supposed theorem. Hence, the random variable $g := \sum_{n=1}^\infty 2^{-n} Y_n$ is also $0$ almost surely.
But $g = f + X_1$, which implies $X_1 \equiv 0$ almost surely. After renormalization, this means $X_1 \equiv \mu$ a.s. But it is clear that one can find examples of sequences $(X_n)_n$ satisfying your assumptions such that $X_1 \equiv \mu$ does not hold almost surely.