# Calculation of a 'double' sum

Let $n \in N$ and $q\geq 2$.

I am trying to calculate the following sum: $$\sum_{i=0}^{\sqrt n/2}\sum_{j= i \sqrt n }^{(i+1)\sqrt n}\frac{(-1)^q2^q(\frac{n}{2}-j)^q}{(n-j)!j!}$$

Any help will be appreciated.

Thank you.

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Is there a question? – J. M. Apr 17 '12 at 3:23
How are you interpreting your bounds on the sums when $\frac{n}{2}$ and $\sqrt{n}$ are not whole numbers? – brc Apr 17 '12 at 3:41
j can go to (n^3/2)/2, so (n-j)! will at some points refer to factorials of negative numbers. Expected? – Wonder Apr 17 '12 at 4:10
Yes. We can 'stop' the sum with negative factorials. Can think that they are 0. – David Apr 17 '12 at 4:20
Shouldn't you be using the floor or ceiling function in the limits? As brc is pointing out, how would you summ up to $\sqrt 3/2$? – Pedro Tamaroff Apr 17 '12 at 5:16