Prove that $S_i<1$ for all $i$ by induction We define $x_1=\frac12$, and $x_n=(1-\frac{3}{2n})\cdot x_{n-1}$
Then we define $S_i=x_1+x_2+\cdots+x_i$ 
Prove that $S_i<1$ for all $i$
I can see that $x_n=(1-\frac{3}{2n})\cdot (1-\frac{3}{2n-2})\cdots (1-\frac 34)\cdot \frac12$
But what then? I suppose my observation isn't useful though...

I think it will help to observe,
\begin{align*}
x_n &= \frac{1}{2} \cdot \frac{2n - 3}{2n} \cdot \frac{2n - 5}{2n - 2} \cdot \ldots \cdot \frac{1}{4} \\
&= \frac{(2n - 3)(2n - 5) \ldots 1}{(2n)(2n - 2)(2n - 4) \ldots 2} \\
&= \frac{1}{2^n n!}(2n - 3)(2n - 5) \ldots 1 \\
&= \frac{1}{2^n n!} \cdot \frac{(2n - 2)(2n - 3)(2n - 4) \ldots 1}{(2n - 2)(2n - 4) \ldots 2} \\
&= \frac{1}{2^n n!} \cdot \frac{(2n - 2)!}{2^{n-1}(n-1)!} \\
&= \frac{(2n - 2)!}{2 \cdot 4^{n-1} n ((n - 1)!)^2} \\
&= \frac{\binom{2n - 2}{n - 1}}{n 2^{2n-1}} \\
&= \frac{C(n - 1)}{2^{2n - 1}},
\end{align*}
where $C(n)$ is the $n$th Catalan number. According to Wikipedia, the Catalan numbers have the following generating function:
$$c(x) = \frac{1-\sqrt{1-4x}}{2x}.$$
With any luck, we should have,
$$\sum_{n=1}^\infty x_n = \frac{1}{2}c\left(\frac{1}{4}\right) = 1,$$
provided the Maclaurin Series converges (removing the removable discontinuity at $0$) where we want. Unfortunately, this value lies right on the boundary of the radius of convergence, but it does mean that this sum equalling $1$ is equivalent to  the Maclaurin series of $\sqrt{1 - x}$ converging to $0$ at $x = 1$. It seems like a more elementary problem, but I don't personally have a proof that it's true.
