# Sums of Double series where $a_{m,n}=\dfrac {m-n} {2^{m+n}}\dfrac {\left( m+n-1\right) !} {m!n!}$

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.

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Here is a possible approach:

$$\frac{1}{(1-x)^{m+1}} = \sum_{n=0}^{\infty} \binom{m+n}{n} x^n$$

Write $\displaystyle a_{m,n}$ as $\displaystyle \frac{m}{(m+n)2^{m+n}} \binom{m+n}{n}$ - $\displaystyle \frac{n}{(m+n)2^{m+n}} \binom{m+n}{n}$

For the first term, multiply by $x^m$ and integrate between $0$ and $1/2$.

For the second terms, differentiate first, the multiply by $x^{m+1}$ and integrate.

When you sum over $m$, you can move the sum into the integral and can possibly simplify it (I haven't worked out the details...)

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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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That's quite interesting i had n't thought of it like that. Can we conclude from this that the question is ill-formed ? –  Hardy Mar 12 '12 at 0:16
Once you specify $a_{0,0} = 0$, it seems to be another example similar to math.stackexchange.com/questions/118945 –  Robert Israel Mar 12 '12 at 0:54