Generalized hypergeometric functions V

Nice infnite series involving binomial coefficients

Today we show the proof of this nice series posted by @infseriesbot \[\sum_{k=0}^{\infty} \frac{\binom{a}{k}\binom{b}{k}\binom{c}{k}}{\binom{a+b+c}{k}} = \frac{(a+b)!(b+c)!(c+a)!}{a!b!c!(a+b+c)!}\] The proof will rely on some properties of the rising factorial and the generalized hypergeometric function.

Proof

First, we need to recall some properties of the binomial coefficient and the rising factorial: The following formula \[ \binom{v}{m} = \frac{(v-m+1)_{m}}{m!} \tag{1}\] relates the binomial coefficient to the Pochhammer polynomial (rising factorial) where \[(x)_{n}= x(x+1)(x+2)\cdots(x+n-1)\] is the definition for the rising factorial.

It is easy to show this. Just note that \[\binom{v}{m} = \frac{v!}{m!(v-m)!} = \frac{(v-m+1)(v-m+2)\cdots(v-1)\cdot v}{m!} \] The numerator in the right hand side is the pochhammer polynomial $(v-m+1)_{n}$ also called the falling factorial.

Pochhammer polynomials obey the reflection formula: \[(-x)_{n} = (-1)^n(x-n+1)_{n} \tag{2}\] Note that, by definition \[ (x-n+1)_{n} = \underbrace{(x-n-1)(x-n)(x-n+1)\cdots (x-1)\cdot x }_{\textrm{n}-terms} \] Hence \[ (-1)^n(x-n+1)_{n} =(-x)(-x+1)\cdots (-x+n-1) = (-x)_{n} \] Hence, back to our series: \begin{align*} S = \sum_{k=0}^{\infty} \frac{\binom{a}{k}\binom{b}{k}\binom{c}{k}}{\binom{a+b+c}{k}} = &\sum_{k=0}^{\infty} \frac{\frac{(a-k+1)_{k}}{k!}\frac{(b-k+1)_{k}}{k!}\frac{(c-k+1)_{k}}{k!} }{\frac{(a+b+c-k+1)_{k}}{k!}} \quad (\textrm{from } (1))\\ =&\sum_{k=0}^{\infty} \frac{(a-k+1)_{k}(b-k+1)_{k}(c-k+1)_{k} }{k!k!(a+b+c-k+1)_{k}}\\ =& \sum_{k=0}^{\infty} \frac{(-1)^k(-a)_{k}(-1)^k(-b)_{k}(-1)^k(-c)_{k} }{k!k!(-1)^k(-a-b-c)_{k}} \quad (\textrm{from } (2))\\ =& \sum_{k=0}^{\infty} \frac{(-a)_{k}(-b)_{k}(-c)_{k} }{(1)_{k}(-a-b-c)_{k}}\frac{1}{k!} \quad ((1)_{k} = k!)\\ =& {{}_{3}F_{2}}\left({-a,-b,-c\atop 1,-a-b-c};1\right) \end{align*} Now we will make use of the Saalschütz theorem (proof in the Appendix): \[{{}_{3}F_{2}}\left({-n,e,f\atop g,h};1\right)=\frac{{\left(g-e\right)_{n}}{% \left(g-f\right)_{n}}}{{\left(g\right)_{n}}{\left(g-e-f\right)_{n}}},\] where $g+h = e+f+1-n \quad n=0,1,2,..$.

Note that in our case $-n = -a$ and $g=1$ so the series satisfy the hypothesis. This kind of series are called Saalschutzian.

Therefore \[S = {{}_{3}F_{2}}\left({-a,-b,-c\atop 1,-a-b-c};1\right) = \frac{(1+a)_{c}(1+b)_{c}}{(1)_{c}(1+a+b)_{c}}\] Now recall that Pochhammer polynomials can be expressed as the quotient of two gamma functions: \[(x)_{n} = \frac{\Gamma(n+x)}{\Gamma(x)} \] Hence \[S = {{}_{3}F_{2}}\left({-a,-b,-c\atop 1,-a-b-c};1\right) = \frac{\frac{\Gamma(c+a+1)}{\Gamma(a+1)}\frac{\Gamma(c+b+1)}{\Gamma(b+1)}}{\Gamma(c+1)\frac{\Gamma(c+a+b+1)}{\Gamma(a+b+1)}} = \frac{\Gamma(c+a+1)\Gamma(c+b+1)\Gamma(a+b+1)}{\Gamma(c+1)\Gamma(a+1)\Gamma(b+1)\Gamma(a+b+c+1)}\] Using $ \Gamma(n+1) = n!$, we can conclude \[ \boxed{ \sum_{k=0}^{\infty} \frac{\binom{a}{k}\binom{b}{k}\binom{c}{k}}{\binom{a+b+c}{k}} = \frac{(a+b)!(b+c)!(c+a)!}{a!b!c!(a+b+c)!} \quad a,b,c\in \mathbb{N}_{0}}\]

Appendix.

Now, we will provide a proof for the Saalschütz theorem: \[{{}_{3}F_{2}}\left({-n,e,f\atop g,h};1\right)=\frac{{\left(g-e\right)_{n}}{% \left(g-f\right)_{n}}}{{\left(g\right)_{n}}{\left(g-e-f\right)_{n}}},\] where $g+h = e+f+1-n \quad n=0,1,2,..$

Proof

Consider the Euler's identity for ${}_{2}F_{1}$: \[ {}_{2}F_{1}(g-e,g-f;g;z) = (1-z)^{e+f-g}{}_{2}F_{1}(e,f;g;z)\] Note that the left hand side is: \[{}_{2}F_{1}(g-e,g-f;g;z) = \sum_{n=0}^{\infty} \frac{(g-e)_{n}(g-f)_{n}}{(g)_{n}}\frac{z^n}{n!} \tag{1} \] while for the right hand side, if we expand $\displaystyle (1-z)^{e+f-g}$: \[(1-z)^{e+f-g} = \sum_{k=0}^{\infty}\binom{e+f-g}{k}(-1)^kz^k = \sum_{k=0}^{\infty}\frac{(g-e-f)_{k}}{k!}z^k\] Hence \[ (1-z)^{e+f-g}{}_{2}F_{1}(e,f;g;z) = \sum_{j=0}^{\infty} \frac{(e)_{j}(f)_{j}}{(g)_{j}}\frac{z^j}{j!}\sum_{k=0}^{\infty}\frac{(g-e-f)_{k}}{k!}z^k\] By the identity theorem for power series the coefficients of the left hand side should be equal to the coefficients of the right hand side i.e. $n=k+j$ So, if we put $k = n-j$ \begin{align*} \sum_{j=0}^{n} \frac{(e)_{j}(f)_{j}(g-e-f)_{n-j}}{(g)_{j}j!(n-j)!} =& \frac{(g-e-f)_{n}}{n!} \sum_{j=0}^{n} \frac{(e)_{j}(f)_{j}(-n)_{j}}{j!(g)_{j}(1-g+e+f-n)_{j}}\\ =& \frac{(g-e-f)_{n}}{n!} {}_{3}F_{2}(e,f,-n;g,1+e+f-g-n;1) \tag{2} \end{align*} From (1) and (2) we can conclude \[{{}_{3}F_{2}}\left({-n,e,f\atop g,h};1\right)=\frac{{\left(g-e\right)_{n}}{% \left(g-f\right)_{n}}}{{\left(g\right)_{n}}{\left(g-e-f\right)_{n}}},\] where $g+h = e+f+1-n \quad n=0,1,2,..$