Let's define
$$
\beta_n \doteq \sum_{i\le (n-1)/2 } \binom{n-(i+1)}{i} (-1)^i \frac{1}{ (2i+1) 2^{2i+1} }.
$$
The following problem is equivalent to proving that $S=0$:
prove that the sequence $\beta_n$ satisfies the recursion
$$
\beta_{n+1} = \frac{2n+1}{2n+2} \beta_n +\frac{1}{(n+1) 2^{n+1}}.
$$
Similar with $S=0$, numerical computations suggest that this statement is true. Unfortunately, I didn't see a straightforward way to prove it.

Below is one way to think about the problem, which led to the above reformulation.

**The connection between the above problem and $S=0$.**

Using the notation developed in the previous answer, let's define
$$
F(m,n) = \sum_{k=0}^m (-1)^k \binom{m}{k} \binom{2(n+k)}{n+k} \frac{1}{2^{2(n+k)}} \sum_{l=1}^{k+n}
\frac{2^l}{l \binom{2l}{l} },
$$
and
$$
f(n)= F(0,n)= \binom{2n}{n} \frac{1}{2^{2n}}
\sum_{l=1}^n \frac{2^l}{l \binom{2l}{l} }.
$$
The statement $S=0$ is the same as $F(m,m)= 0$. Note that $F$ satisfies
$$
F(m,n) = \frac{1}{2} F(m-1,n) - \frac{1}{2}F(m-1,n+1) ~~~~~~\text{(r1)}
$$
Define the difference operator $D(x_1,x_2) = (x_1 - x_2)/2.$ (r1) in terms of $D$
is
$$
F(m,n) = D( F(m-1,n), F(m-1,n+1) ).
$$
Define $D^k$ by iterating $D$:
$$
D^n(x_1,x_2,x_3,\ldots,x_{n+1}) = D( D^{n-1}(x_1,x_2,x_3,\ldots,x_{n}), D^{n-1}(x_2,x_3,\ldots,x_{n+1} ))
$$
Iterating (r1) gives

$$
F(m,n) = D^m( f(n),f(n+1),f(n+2), f(n+3),\cdots,f(n+m)).
$$

In particular:
$$
F(m,m) = D^m( f(m),f(m+1),f(m+2), f(m+3),\cdots,f(m+m)).
$$

Define
${\mathcal D}:{\mathbb R}^\infty\rightarrow {\mathbb R}^\infty$ as follows:
the $i^{th}$ component of ${\mathcal D}(x_{1}^\infty)$ is
$$D^n(x_n,x_{n+1},x_{n+2},\ldots,x_{2n}).$$

We can restate our original problem as follows:
show that $(f(1),f(2),f(3),...,f(n),...)$ is in the kernel
of ${\mathcal D}$.

Because we are looking for a zero of this operator,
the $1/2$ in the definition of $D$ is not important; thus let us assume that $D(x_1,x_2)$ is simply
$x_1 -x_2$.

Note that
$D^{n}(f(n),f(n+1),...,f(2n)) =0$
is the same as
$$
D^{n-1}(f(n),f(n+1),f(n+2),...,f(2n-1)) = D^{n-1}(f(n+1),f(n+2),f(n+3),...,f(2n)).
$$
A numerical computation reveals that these discrete derivatives equal $\frac{1}{(2n-1)2^{2n-1}}$.
One can go back from these values
to an element
of the kernel of
${\mathcal D}$
by inverting each $D$ in the above display.
A bit of computation in this direction yields
the vector $\beta$ in the first display.
By its construction $\beta$ is in the kernel of ${\mathcal D}$.
Thus if one can prove that $f$ equals $\beta$ then we are done.

Finally, using its definition, we see that $f$ satisfies:
$$
f(n+1) = \frac{2n+1}{2n+2} f(n) + \frac{1}{(n+1)2^{n+1}}, ~~~ f(1) = 1/2.
$$
These relations determine $f$ and thus we can take them as $f$'s definition.
Thus to verify $f=\beta$ it is enough to show that $\beta$ satisfies this recursion.

computerhas written more papers than you have” (directed at some upstart faculty member, perhaps)? $\endgroup$ – LSpice Mar 4 '10 at 16:18