1. limits in double sum

Could you explain me the change in the limits of this double sum?
$\sum_{a=0}^n\binom{n}{a}(1+x(1-y))^{n+a}x^a(y+x(1-y))^a\sum_{b=0}^{n+a}\binom{n+a}{b}k^b=$

$=\sum_{b=0}^{2n}\sum_{a=max(0,b-n)}^n\binom{n}{a}\binom{n+a}{b}(1+x(1-y))^{n+a}x^a(y+x(1-y))^ak^b$

Is there any property about this change and how can I understand the correct change in a similar case?

2. Re: limits in double sum

Denote by $f(a,b)$ what is summed. We have to show that $\sum_{a=0}^n\sum_{b=0}^{n+a} f(a,b) =\sum_{b=0}^{2n}\sum_{a=\max (0,b-n)}^nf(a,b)$. We have

\begin{align*}\sum_{b=0}^{2n}\sum_{a=\max (0,b-n)}^nf(a,b) &=\sum_{b=0}^{n}\sum_{a=\max (0,b-n)}^nf(a,b)+\sum_{b=n+1}^{2n}\sum_{a=\max (0,b-n)}^nf(a,b) \\&= \sum_{b=0}^{n}\sum_{a=0}^nf(a,b)+\sum_{b=n+1}^{2n} \sum_{a=b-n}^nf(a,b)\end{align*}
and $\sum_{a=0}^n\sum_{b=0}^{n+a} f(a,b) =\sum_{a=0}^n\sum_{b=0}^n f(a,b)+\sum_{a=0}^n\sum_{b=n+1}^{n+a} f(a,b)$. Hence you only have to show that $\sum_{b=n+1}^{2n}\sum_{a=b-n}^nf(a,b) =\sum_{a=0}^n\sum_{b=n+1}^{n+a} f(a,b)$. To see that, we notice that the set in which $a$ and $b$ lie is a triangle.

3. Re: limits in double sum

Originally Posted by girdav
... we notice that the set in which $a$ and $b$ lie is a triangle.
Could you explain me what you mean?

4. Re: limits in double sum

For the first sum, we sum for $(a,b) = (1,n+1),\ldots (n,n+1), (2,2+n),\ldots, (n,n+2),\ldots, (n,2n)$ and if your draw these points on the upper half plane you will get a triangle.