2 corrected several typos

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 $$\beta_{n+1} = \frac{2n+1}{2n+2} \beta_n +\frac{1}{n 2^n}\frac{1}{(n+1) 2^{n+1}}.$$ 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 $$F(m,n) = \sum_{k=0}^m (-1)^k \binom{m}{k} \binom{2(n+k)}{d+k} binom{2(n+k)}{n+k} \frac{1}{2^{2(n+k)}} \sum_{l=1}^{k+n} \frac{2^l}{l \binom{2l}{l} },$$ 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 $\frac{1}{(2i+1)2^{2i-1}}$. \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. 1 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 $$\beta_{n+1} = \frac{2n+1}{2n+2} \beta_n +\frac{1}{n 2^n}.$$ 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 $$F(m,n) = \sum_{k=0}^m (-1)^k \binom{m}{k} \binom{2(n+k)}{d+k} \frac{1}{2^{2(n+k)}} \sum_{l=1}^{k+n} \frac{2^l}{l \binom{2l}{l} },$$ 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$\frac{1}{(2i+1)2^{2i-1}}$. 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.