4

I'm trying to find some estimates for some PDE I'm working on. I'd like to estimate the sum $$\sum_{|\alpha|\leq p} \binom{p}{|\alpha|},$$ where $\alpha\in\mathbb N_0^n$ and $\mathbb N_0=\mathbb N\cup\{0\}$. I know, $$\#\{\alpha\in\mathbb N_0^n: |\alpha|\leq p\}=\binom{p+n}{p},$$ hence the following inequality holds, $$\sum_{|\alpha|\leq p} \binom{p}{|\alpha|}\leq \binom{p+n}{p}\max_{|\alpha|\leq p}\binom{p+n}{p}.$$ But I'm wondering if there is a more elegant estimate for the above sum. Can anyone help me?

PtF
  • 9,655
  • 1
    What is $#{\alpha \in \mathbb N^n_0:\ |\alpha|=p}$? Maybe you could transform the sum then? – daw Aug 04 '14 at 19:29
  • The $\alpha$s for which $|\alpha|=k$ are precisely the weak $n$-part integer compositions of $k$. There is an explicit formula for counting these explained on Wikipedia here. Hence the sum is $$(\circ)=\sum_{k=0}^p\binom{p}{k}\binom{k+n-1}{n-1}.$$ WolframAlpha says this equals the hypergeometric function ${}_2F_1(n,-p;1;-1)$. – anon Aug 04 '14 at 19:33
  • $#$ means the cardinality.. – PtF Aug 04 '14 at 20:01

0 Answers0