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 $\Gamma(z+1)=z\Gamma(z)$.



Help with a double sum, please
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


