<s>[INST] Sums of Double series where $a_{m,n}=\frac {m-n} {2^{m+n}}\frac {\left( m+n-1\right) !} {m!n!}$ I am trying to show if $$a_{m,n}=\dfrac {m-n} {2^{m+n}}\dfrac {\left( m+n-1\right) !} {m!n!}$$ such that $$\left( m,n>0\right) $$ that $\sum _{m=0}^{\infty }\left( \sum _{n=0}^{\infty }a_{m,n}\right) =-1$, $\sum _{n=0}^{\infty }\left(\sum _{m=0}^{\infty } a_{m,n}\right) =1$.
Now we observe that $a_{m,0}=2^{-m}$,$a_{0,n}=-2^{-n}$ and $a_{0,0}=0$.
We can rewrite the first series as 
that $\sum _{m=0}^{\infty }\left( \sum _{n=0}^{\infty }a_{m,n}\right)$.
$\Rightarrow \sum _{m=0}^{\infty }\left(\dfrac {1} {2^{m}}+\dfrac {\left( m-1\right) } {2^{m+1}}+\dfrac {\left( m-2\right) \left( m+1\right) } {2^{m+2}2!}+\dfrac {\left( m-3\right) \left( m+1\right) \left( m+2\right) } {2^{m+3}3!}+\ldots\right)$
$\Rightarrow \sum _{m=0}^{\infty }\dfrac {1} {2^{m}}\left(1+\dfrac {\left( m-1\right) } {2^{1}}+\dfrac {\left( m-2\right) \left( m+1\right) } {2^{2}2!}+\dfrac {\left( m-3\right) \left( m+1\right) \left( m+2\right) } {2^{3}3!}+\ldots\right)$
I can n't recognize the expression inside the brackets, any help of how to proceed forward would be much appreciated.
 [/INST] It's not clear how to define $a_{0,0}$, since $(-1)!$ is undefined.  But if you take $a_{0,0}$ to be $0$ you should have $\sum_{n=1}^\infty a_{0,n} = -1$ while $\sum_{n=0}^\infty a_{m,n} = 0$ otherwise.  Note that $$a_{m,n} = 2^{-m-n} \left({m+n-1 \choose n} - {m+n-1 \choose m} \right)$$
and 
$$ { m+n-1\choose n} = (-1)^n {-m \choose n}$$
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