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We know that $B(p,q)=\Gamma(p)\Gamma(q)/\Gamma(p+q)$ where $p, q>0$, and $B(p,q)$ is related to binomial coefficients if one of $p,q$ is an integer. I want to prove the following identity.

$$\frac{\Gamma(b)}{\Gamma(b+2n+3)}\sum_{k=0}^n \frac{(-1)^{k+1}\Gamma(b+n+k+2)}{(n-k+1)!(k+1)!\Gamma(b+k+1)}=\frac{1}{(n+1)!(n+2)!}\left[(-1)^{n+1}B(b,n+2)-B(b+n+1,n+2)\right],$$

where all variables except $b$ are nonnegative integers, and $b=p/2+q$ for nonnegative integers $p,q.$ In fact, the value of $b$ is not essential here, as long as $b\geq 0$.

I tried to convert the above expression into an identity about binomial coefficients but still could not figure it out.

Note that for the special case where $n=0$, it can be verified that the above identity holds.

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  • $\begingroup$ The general term of your series seems to be of the form $\dfrac{(x+y+z)!}{x!~y!~z!}$ , which has trinomial coefficient written all over it. $\endgroup$ Commented May 10, 2015 at 16:08
  • $\begingroup$ You are right, but I do not see essential benefits out of it. For example, we can write $$\frac{(x+y+z)!}{x!y!z!}=\frac{(x+y+z)!}{x!(y+z)!}\frac{(y+z)!}{y!z!},$$ which is a product of two binomial coefficients. Moreover, since $b$ may not be an integer, it would be troublesome to shift to trinomial or multinomial coefficient. $\endgroup$ Commented May 10, 2015 at 16:49

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Suppose we are interested in the value of $$S_b(n) = \sum_{k=0}^n \frac{(-1)^{k+1}}{(n-k+1)!(k+1)!} \frac{\Gamma(b+n+k+2)}{\Gamma(b+k+1)}.$$

This is $$\sum_{k=0}^n \frac{(-1)^{k+1}}{(n-k+1)!(k+1)!} (n+1)! {b+n+k+1\choose n+1} \\ = \frac{1}{n+2} \sum_{k=0}^n \frac{(-1)^{k+1} (n+2)!}{(n-k+1)!(k+1)!} {b+n+k+1\choose n+1} \\ = \frac{1}{n+2} \sum_{k=0}^n (-1)^{k+1} {n+2\choose k+1} {b+n+k+1\choose n+1} \\ = \frac{1}{n+2} \sum_{k=1}^{n+1} (-1)^k {n+2\choose k} {b+n+k\choose n+1}.$$

This becomes $$-\frac{1}{n+2} {n+b\choose n+1} - \frac{1}{n+2} (-1)^n {2n+b+2\choose n+1} \\ + \frac{1}{n+2} \sum_{k=0}^{n+2} (-1)^k {n+2\choose k} {b+n+k\choose n+1}.$$

To evaluate the sum we introduce $${b+n+k\choose n+1} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{b+n+k}}{z^{n+2}} \; dz.$$

This yields for the sum $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{b+n}}{z^{n+2}} \sum_{k=0}^{n+2} (-1)^k {n+2\choose k} (1+z)^{k} \; dz \\ = \frac{(-1)^{n+2}}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{b+n}}{z^{n+2}} (-1+1+z)^{n+2} \; dz \\ = \frac{(-1)^{n}}{2\pi i} \int_{|z|=\epsilon} (1+z)^{b+n} \; dz = 0.$$

It follows that $$S_b(n) = -\frac{1}{n+2} {n+b\choose n+1} - \frac{1}{n+2} (-1)^n {2n+b+2\choose n+1}.$$

Converting this into the gamma function format is pure algebra and left as an exercise to the reader.

Remark. We use $$(1+z)^{b+n} = \exp((b+n)\log(1+z))$$ when $b$ is not an integer where the branch cut of the logarithm is on the negative real axis so that the branch point is at $z=-1.$ This ensures that we have analyticity in a disk enclosing the origin.

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  • $\begingroup$ Alles klar! An integral representation for binomial coefficients (similar to the Euler integral for beta function) is what I am looking for! Unfortunately I spent two days and was not smart enough to connect the definition of binomial coefficient to residue theorem then. Thanks for your explanation. $\endgroup$ Commented May 12, 2015 at 6:36

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