Help with a double sum, please Here is a double series I have been having trouble evaluating:
$$S=\sum_{k=0}^{m}\sum_{j=0}^{k+m-1}(-1)^{k}{m \choose k}\frac{[2(k+m)]!}{(k+m)!^{2}}\frac{(k-j+m)!^{2}}{(k-j+m)[2(k-j+m)]!}\frac{1}{2^{k+j+m+1}}\text{.}$$
I am confident that $S=0$ for any $m>0$. In fact, I have no doubt. I have done lots of algebraic manipulation, attempted to "convert" it to a hypergeometric series, check tables (Gradshteyn and Ryzhik), etc., but I have not been able to get it into a form from which I can prove zero equivalence.
Here is another form of the sum (well, I hope at least) that might be easier to work with:
$$S=\sum_{k=0}^{m}\sum_{j=0}^{k+m-1}(-1)^{k}\frac{m!}{k!(m-k)!}\frac{(k+m-1-j)!}{(k+m)!}\frac{(k+m-1/2)!}{(k+m-1/2-j)!}\frac{1}{2^{k-j+m+1}}\text{.}$$
I have read Concrete Mathematics and $A=B$, and looked at Gosper's and Zeilberger's work for some hints, but no cigar.
Note: $0!=1$ and $n!=n(n-1)!$ for $n\in\mathbb{N}\cup\{0\}$. For $n\in\mathbb{R}^+$, $n!=n\Gamma(n)$ where $\Gamma\colon\: \mathbb{C}\to\mathbb{C}$ and, for $\Re z>0$ and $z\notin\mathbb{Z}^{-}$, $$\Gamma\colon\: z\mapsto  \int_0^\infty t^{z-1}\mathrm{e}^{-t}\,\mathrm{d}t\text{.}$$
which can be analytically extended to $\mathbb{C}$ via the recurrence $\Gamma(z+1)=z\Gamma(z)$.
 A: Shall we try teamwork?  Please feel free to edit this post if you have simplifications.
The original sum may be re-expressed as
$$ \frac{1}{2^{2m+1}} \sum_{k=0}^m (-1)^k \binom{m}{k} \binom{2(k+m)}{k+m} \frac{1}{2^{2k}} \sum_{j=0}^{k+m-1} \frac{2^{k+m-j}}{(k+m-j) \binom{2(k+m-j)}{k+m-j}}. $$
If we're trying to prove this is 0, we may drop the fraction out front.  Also, change variables from $j$ to $\ell=k+m-j$:
$$ \sum_{k=0}^m \left( -\frac14 \right)^k \binom{m}{k} \binom{2(k+m)}{k+m} \sum_{\ell=1}^{k+m} \frac{2^\ell}{\ell \binom{2\ell}{\ell}}. $$
At this point, my idea was to change the order of summation based on
$$ \sum_{k=0}^m \sum_{\ell=1}^{k+m} \Diamond = \sum_{\ell=1}^m \sum_{k=0}^m \Diamond + \sum_{\ell=m+1}^{2m} \sum_{k=\ell-m}^m \Diamond, $$
but I can't get quite it to work out.  The first sum simplifies, but the second sum I can't do much with.
Any ideas?
A: Some observations.
Define 
$$T(N)=\binom{2N}{N}\sum_{j=0}^{N-1}\left[\binom{2(N-j)}{N-j} \cdot (N-j) \cdot 2^{(N+j+1)}\right]^{-1}.$$
Then 
$$S(m)=\sum_{k=0}^m (-1)^k \binom{m}{k} T(m+k) .$$
Experimentation shows that $T$ satisfies the recursion
$$T(n)=T(n-1) - \frac{1}{12}T(n-2),$$
though I don't know how to prove that.
For any $F$ satisfying such a recursion
$$F(n) = F(n-1) - c \cdot F(n-2)$$
we have
$$\sum_{k=0}^m (-1)^k \binom{m}{k} F(m+k) = c^m F(0),$$
which is probably easy to prove; and $T(0)=0$.
A: Olivier Gerard just told me about this wonderful website!
Regarding the question it can be done in one nano-second using the Maple 
package 
http://www.math.rutgers.edu/~zeilberg/tokhniot/MultiZeilberger
accopmaying my article with Moa Apagodu
http://www.math.rutgers.edu/~zeilberg/mamarim/mamarimhtml/multiZ.html
Here is the command:
F:=(-1)^k*binomial(m,k)*(k+m-1-j)!/(k+m)!*simplify((k+m-1/2)!/(k+m-1/2-j)!)/2^(k-j):  lprint(MulZeil(F,[j,k],m,M,{})[1]);
and here is the output:
-1/4*(2*m+1)/(m+1)+M
(Note that I had to divide the summand by 1/2^(m+1) if you don't you get FAIL,
the prgram does not like extraneous factors)
Translated to humaneze we have that (my S(m) is hte original S(m) times 2^(m+1))
S(m+1)=(2m+1)/(m+1)S(m)
Since S(1)=0 (check!) This is a completely rigorous proof.
P.S. The proof can be gotten by finding the so-called multi-certificate
lprint(MulZeil(F,[j,k],m,M,{})[2]);
-Doron Zeilbeger
A: has2's beta satisfies 
$$4(n + 2)\beta_{n + 2} - 2(3n + 4)\beta_{n + 1} + (2n + 1)\beta_n= 0$$
(use any guessing package) which might be easier to use. 
A: Better than reading A=B, you should download and learn to use their Mathematica package. I've used it successfully in the past to obtain automagic proofs of similar identities.
A: Well, if algebra doesn't help, let's try good old complex analysis. Put $u=k-j+m$, $s=k+m$ and rewrite the sum as
$$
\sum_{u\ge 1}2^u\frac{u!(u-1)!}{(2u)!}[x^m]\sum_{s\ge u}2^{-2s}{2s\choose s}x^s(1-\frac 1x)^m
$$
where $[x^m]F(x)$ is the $m$-th Laurent coefficient of $F$ at $0$.
Now we can at least recognize the coefficients. The sum in $s$ is just the truncated Taylor sum for $\frac 1{\sqrt{1-x}}$ and $2^u\frac{u!(u-1)!}{(2u)!}=\int_0^1{[2t(1-t)]^u}\frac{dt}{t}$. Recalling that the truncation of analytic functions to high frequences is just $z^uP_+z^{-u}$ where $P_+$ is the Cauchy integral, and that the coefficient at the $m$-th power can be obtained by integration against $z^{-m}$ over a circle, we can write this monster as
$$
\int_0^1 \frac{dt}{t}\sum_{u\ge 1}[2t(1-t)]^u\oint\oint \frac{z^u z^{-2m}(z-1)^m}{\zeta^u\sqrt{1-\zeta}(1-\frac z\zeta)}dm(z)dm(\zeta)
$$
with circular integral taken over the circles of radii less than $1$ with the radius for $z$ smaller than that for $\zeta$ ($m$ is the averaging measure here, so the integrals are just the averages over the corresponding circles).
Now, summing over $u$, we get 
$$
\int_0^1 \frac{dt}{t}\oint\oint\left(\frac{1}{1-2t(1-t)\frac z\zeta}-1\right) \frac{z^{-2m}(z-1)^m}{\sqrt{1-\zeta}(1-\frac z\zeta)}dm(z)dm(\zeta)
$$
Now $(\frac{1}{1-pw}-1)\frac 1{1-w}=\frac p{1-p}(\frac1{1-w}-\frac1{1-pw})$
Thus, using the Cauchy formula again and integrating over $\zeta$, we convert it into
$$
\int_0^1 \frac{2(1-t)dt}{1-2t(1-t)}\oint\left(\frac{1}{\sqrt{1-z}}-\frac{1}{\sqrt{1-2t(1-t)z}}\right) {z^{-2m}(z-1)^m}dm(z)
$$
The integral in $t$ is an elementary function of $z$ analytic near the origin (have a nice CAS!)  The claim that the integral is $0$ for all $m$ is equivalent to the claim that after change of variable $w=\frac z{\sqrt {1-z}}$ all the Taylor coefficients of the new integrand in $w$ with even indices are $0$, i.e., the new integrand is an odd function in $w$ (have more nice CAS!). Whether true or false, it is verifiable now. So, I'll stop here -:).  
Edit: It is true. After some moderately tedious computations, it boils down to the fact that $\operatorname{arctan}\sqrt{1-z}-\frac\pi 4$ is an odd function of $w=\frac{z}{\sqrt{1-z}}$, which, believe it or not, is correct. I think you have already checked it using those cute CAS programs, which I haven't on my old laptop, so I'm not posting the details.
Of course, the challenge to find a combinatorial interpretation of this formula still remains.   
A: 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.
