# Identity involving Pochhammer symbol

I came across the following identity in my research: $$\sum_{m=0}^s \frac{(-1)^m (a+2m)}{m!(s-m)! (a+m)_{s+1}}=\delta_{s,0}$$

where $$(a)_n= a(a+1)\cdots (a+n-1)$$ is the Pochhammer symbol. One can experiment with Mathematica and check that this identity is true for $$s=0,1,2....$$ but I don't see an easy way to prove it. I am not very familiar with Pochhammer symbol identities and I am hoping someone can give me some hint.

Thanks :)

• I am genuinely wondering who and why downvoted this question, this looks like a perfectly normal MO question? Jan 29 at 12:14
• @Nemo what the hell do you mean, if the question someone encountered in his research is not appropriate for MO, then what is? Jan 29 at 13:14

Another option is the Wilf-Zeilberger (WZ) method via Zeilberger's algorithm.

Start with the "seed" function (your summand) $$F(s,m):=\frac{(-1)^m(a+2m)\binom{s}m}{\binom{a+m+s}{s+1}s!(s+1)!}.$$ The algorithm generates a "companion" function $$G(s,m):=\frac{(-1)^{m+1}\binom{s-1}{m-1}}{\binom{a+m+s-1}ss!^2}.$$ Then, we ask a computer algebra software to verify that $$F(s,m)=G(s,m+1)-G(s,m)$$: one easier way is to divide both sides by $$F(s,m)$$ because the task would reduce to testing equality between two rational (or polynomial) expressions.

Next, sum over the integers $$m\in\{0,1,\dots,s\}$$ to obtain: $$\sum_{m=0}^sF(s,m)$$ which is the LHS of your claim, while $$\sum_{m=0}^s(G(s,m+1)-G(s,m))$$ telescopes to $$G(s,s+1)-G(s,0)$$. A quick check shows $$G(s,s+1)=0$$ and that $$-G(s,0)=0$$ unless $$s=0$$, in which case $$-G(s,0)=1$$.

In summary, we arrive at $$\sum_{m=0}^sF(s,m)=\delta_0(s)$$.

• +1. Since the OP expressed unfamiliarity with this subject, it's worth mentioning that the WZ method can prove a large class of identities automatically, without "human cleverness." Learning the WZ method therefore provides a lot of "bang for the buck"; with only a modest investment of time, one greatly increases the number of identities one can prove without having to consult an expert. Jan 30 at 14:04
• @TimothyChow: you are considerate, thank you for the comments. Jan 30 at 14:41
• Wow this is amazing! I've never actually heard of WZ method! Thanks a lot for the answer @T.Amdeberhan
– XYX
Feb 1 at 4:44
• And thanks for the comment! @TimothyChow
– XYX
Feb 1 at 4:45

You had several nice answers proving the identity from scratch. Let me point out that what you call "Pochhammer symbols identities" is also known as hypergeometric identities. If you encounter an identity of this type, it is usually well-known and you can look it up in any decent book on special functions. All you need to do is write it in standard notation, which in this case is $${}_3F_2\left(\begin{matrix}-s,a/2+1,a\\a/2,a+1+s\end{matrix};1\right)=\delta_{0,s}.$$ This is indeed a special case of Dixon's identity (see e.g. 16.4.4 on https://dlmf.nist.gov/16.4) $${}_3F_2\left(\begin{matrix}a,b,c\\1+a-b,1+a-c\end{matrix};1\right)=\frac{\Gamma(1+a/2)\Gamma(1+a-b)\Gamma(1+a-c)\Gamma(1+a/2-b-c)}{\Gamma(1+a)\Gamma(1+a/2-b)\Gamma(1+a/2-c)\Gamma(1+a-b-c)},$$ subject to convergence conditions. If $$c=-s$$ is a non-positive integer, this reduces to the finite sum identity $${}_3F_2\left(\begin{matrix}a,b,-s\\1+a-b,1+a+s\end{matrix};1\right)=\frac{(1+a)_s(1+a/2-b)_s}{(1+a/2)_s(1+a-b)_s}.$$ When $$b=a/2+1$$, the right-hand side indeed reduces to $$\delta_{0,s}$$.

Knowing this, you can ask questions like

• Is my context related to other contexts where Dixon's identity appears?
• Does my context has a natural generalization where some generalization of Dixon's identity would be needed?

You don't get this type of conceptual insight from e.g. the WZ method. Sorry if I'm ranting but this is one of my pet peeves.

• Though I'm a big fan of the WZ method, I agree fully with your remark about conceptual insight. Jan 30 at 14:07

Case $$s=0$$ is clear, so let further $$s>0$$. We twice use Vandermonde-Chu convolution identity $$\sum_{k=0}^n{x\choose k}{y\choose n-k}={x+y\choose n}.$$ Multiplying by $$(a)_{2s+1}=a(a+1)\ldots(a+2s)$$, we get $$\sum_{m=0}^s(-1)^m(a+2m){a+m-1\choose m}{a+2s\choose s-m}=\sum_{m=0}^s(a+2m){-a\choose m}{a+2s\choose s-m}\\ =a{2s\choose s}+2\sum_{m=0}^sm{-a\choose m}{a+2s\choose s-m}=a{2s\choose s}-2a\sum_{m=1}^s{1-a\choose m-1}{a+2s\choose s-m}\\=a{2s\choose s}-2a{2s-1\choose s-1}=0.$$

[Hopefully this doesn’t end up being only a reformulation of the other solutions.]

The $$(-1)^m\frac1{m!(s-m)!}$$ makes me think about discrete derivatives: if you define $$(\Delta f)(x)=f(x+1)-f(x)$$, then $$(\Delta^s f)(x)=\sum_{m=0}^s(-1)^{m-s} {s\choose m}f(x+m).$$ An easy property is that $$\Delta \frac1{(x)_n}=\frac {-n}{(x)_{n+1}}$$. Now we split the factor $$x+2m$$ on top as $$(x+2m)+m$$ and rewrite the expression as \begin{align} & \frac{(-1)^s}{s!}\sum_{m=0}^s(-1)^{m-s}{s\choose m}\frac1{(x+1+m)_s} + \frac{(-1)^s}{(s-1)!} \sum_{m=1}^s (-1)^{s-m}{s-1\choose m-1}\frac1{(x+1+m-1)_{s+1}} \\ &= \frac{(-1)^s}{s!}\left(\Delta^s\frac1{(x+1)_s}\right) + \frac{(-1)^s}{(s-1)!} \left(\Delta^{s-1}\frac1{(x+1)_{s+1}}\right) \\ & = \frac{(-1)^s}{s!}\left(\Delta^s\frac {-s}{(x+1)_{s+1}}\right) + \frac{(-1)^s}{(s-1)!} \left(\Delta^{s-1}\frac1{(x+1)_{s+1}}\right) \\ & = 0\end{align} (We use $$s\ge 1$$ as $$\Delta^{s-1}$$ is not defined for $$s=0$$.)

$$b_m=\frac{ (a+2m)}{m!(s-m)! (a+m)_{s+1}}=\frac{(a+2 m) \Gamma (a+m)}{\Gamma (m+1) \Gamma (-m+s+1) \Gamma (a+m+s+1)}$$

$$S_p=\sum_{m=0}^p(-1)^m \, b_m=(-1)^p\,\frac{(p+1) (a+p+s+1) \Gamma (a+p+1)}{s \Gamma (p+2) \Gamma (s-p) \Gamma (a+p+s+2)}$$

Using algebra, let $$p=ks$$ and expand as a series around $$k=1$$

$$\frac{(p+1) (a+p+s+1) \Gamma (a+p+1)}{s \Gamma (p+2) \Gamma (s-p) \Gamma (a+p+s+2)}=\frac{ \Gamma (a+s+1)}{\Gamma (s+1) \Gamma (a+2 s+1)}(1-k)+O\left((k-1)^2\right)$$