Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

up vote 2 down vote accepted

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...)

share|improve this answer

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 $$ { m+n-1\choose n} = (-1)^n {-m \choose n}$$

share|improve this answer
    
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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.