Probability that sum of random variables is positive does not approach $1$ in the limit Say we have independent random variables $X_i$, with:
$$
X_i = 
\begin{cases}
1,~\text{with probability}~\frac{1}{2}+ 2^{-i},\\
-1,~\text{with probability}~\frac{1}{2}-2^{-i}.
\end{cases}
$$
I'm interested in showing that $\lim_{n\rightarrow \infty}\mathbb{P}[X_1+\dots+X_n > 0]\neq 1$. I've seen an argument for this using the Central Limit Theorem (here, pp. 2-3).
I was wondering: is there perhaps a simpler (more elegant?) argument---perhaps using some more elementary methods, or a well-chosen concentration inequality---that could work well on an audience with less than exhaustive knowledge of statistics?
I tried explicitly spelling out the probability that a majority of the $X_i$'s have value $1$, with the hope of bounding the sum with some inequalities, but with limited success. So am looking for another angle.
 A: If you are not interested in showing that $\mathbb{P}(X_1 + \cdots + X_n > 0)$ actually converges but rather okay with showing
$$ \limsup_{n\to\infty} \mathbb{P}(X_1 + \cdots + X_n > 0) < 1, \tag{*} $$
then we can come up with a quick solution. To this end, we realize the law of $(X_i)_{i\geq 1}$ in a way better suited for analyzing the problem. Suppose that

*

*$Y_i \sim \operatorname{Ber}(\frac{1}{2})$, $i = 1, 2, \ldots,$

*$Z_i \sim \operatorname{Ber}(1-\frac{1}{2^{i-1}})$, $i = 1, 2, \ldots,$ and

*$Y_1, Z_1, Y_2, Z_2, \ldots$ are all independent.

Then we easily find that $(1 - 2 Z_i Y_i)_{i\geq 1} \stackrel{d}= (X_i)_{i\geq 1} $. Using this, the complementary probability can be written as
\begin{align*}
\mathbf{P}\left( X_1 + \cdots + X_n \leq 0 \right)
&= \mathbf{P} \left( \frac{n}{2} \leq Z_1 Y_1 + \cdots + Z_n Y_n \right)
\end{align*}
Let $A_n = \{ Z_1 = 0 \text{ and } Z_i = 1 \text{ for all } i \geq 2 \}$. Then this probability is bounded below by
\begin{align*}
&\mathbf{P} \left( \frac{n}{2} \leq Z_1 Y_1 + \cdots + Z_n Y_n \,\middle|\, A_n \right) \mathbf{P}(A_n) \\
&= \mathbf{P} \left( \frac{n}{2} \leq Y_2 + \cdots + Y_n \right) \prod_{i=2}^{n} \bigl(1 - 2^{-(i-1)} \bigr).
\end{align*}
Taking $\liminf$ as $n\to\infty$, it follows that
$$ \liminf_{n\to\infty} \mathbf{P}\left( X_1 + \cdots + X_n \leq 0 \right)
\geq \frac{1}{2} \prod_{i=2}^{\infty} \bigl(1 - 2^{-(i-1)} \bigr)
> 0 $$
and therefore $\text{(*)}$ is proved.

Remark. Pursuing this direction and performing a more detailed analysis, we get
$$ \lim_{n\to\infty} \mathbf{P}\left( X_1 + \cdots + X_n \leq 0 \right)
= \frac{1}{2}. $$
