Generalized hypergeometric functions IX

Series invoving a quotient of binomial coefficients

We show the proof of this nice result: $$\sum_{k=0}^{n} \frac{{\binom{n}{k}}^3}{\binom{3n}{k}} = \frac{(2n)!^3}{n!^3(3n)!}$$ For the proof we use the Saalschütz theorem for hypergeometric functions.

Proof:

We will transform this series to an hypergeometric series

Note that $$\binom{v}{m} = \frac{(v-m+1)_{m}}{m!}$$ where $$(x)_{m} = x(x+1)(x+2)\cdots (x+m-1)$$ is the Pochhammer polynomial (rising factorial)

Hence $$S= \sum_{k=0}^{n} \frac{{\binom{n}{k}}^3}{\binom{3n}{k}} = \sum_{k=0}^{n} \frac{(n-k+1)_{k}(n-k+1)_{k}(n-k+1)_{k}}{(3n-k+1)_{k}k!k!}$$ Recall that the rising factorial obey the reflection formula: $$(-x)_{k} = (x-k+1)_{k}(-1)^k \tag{*}$$ Therefore $$\begin{align*} S =\sum_{k=0}^{n} \frac{{\binom{n}{k}}^3}{\binom{3n}{k}} =& \sum_{k=0}^{n} \frac{(n-k+1)_{k}(n-k+1)_{k}(n-k+1)_{k}}{(3n-k+1)_{k}k!k!}\\ = & \sum_{k=0}^{n} \frac{(-1)^k(n-k+1)_{k}(-1)^k(n-k+1)_{k}(-1)^k(n-k+1)_{k}}{(-1)^k(3n-k+1)_{k}k!k!}\\ = & \sum_{k=0}^{n} \frac{(-n)_{k}(-n)_{k}(-n)_{k}}{(-3n)_{k} k!k!}\\ = & \sum_{k=0}^{n} \frac{(-n)_{k}(-n)_{k}(-n)_{k}}{(-3n)_{k} (1)_{k}}\frac{1}{k!} \quad \left(k! = (1)_{k} \right) \end{align*}$$ Recall the generalized hypergeometric function $${}_{p}F_{q} \left[ {a_{1},\cdots a_{p}\atop b_{1}\cdots b_{q}}; z\right] = \sum_{k=0}^{\infty} \frac{(a_{1})_{k}(a_{2})_{k}\cdots (a_{p})_{k}}{(b_{1})_{k}(b_{2})_{k}\cdots (b_{q})_{k}} \frac{z^k}{k!}$$ if one of the $a_{i}$ is a negative integer, for example $a_{i} = -n$ with $n\in \mathbb{N}$ we say that the series "terminates" and $${}_{p}F_{q} \left[ {a_{1},\cdots,-n,\cdots, a_{p}\atop b_{1}\cdots b_{q}}; z\right] = \sum_{k=0}^{n} \frac{(a_{1})_{k}(a_{2})_{k}\cdots(-n)_{k} \cdots (a_{p})_{k}}{(b_{1})_{k}(b_{2})_{k}\cdots (b_{q})_{k}} \frac{z^k}{k!}$$ To show this note that $$(-n)_{k} = (-n)(-n+1)\cdots(-n+k-1)$$ Hence $$(-n)_{k} = 0 \quad \textrm{ if } \quad k\geq n+1$$ From this is easy to see that $$S = \sum_{k=0}^{n} \frac{{\binom{n}{k}}^3}{\binom{3n}{k}} = \sum_{k=0}^{n} \frac{(-n)_{k}(-n)_{k}(-n)_{k}}{(-3n)_{k} (1)_{k}}\frac{1}{k!} = {}_{3}F_{2} \left[ {-n, -n, -n \atop -3n, 1}; 1 \right]$$ This series satisfies the Saalschütz theorem: $${{}_{3}F_{2}}\left[{-n,a,b\atop c,d};1\right]=\frac{{\left(c-a\right)_{n}}{% \left(c-b\right)_{n}}}{{\left(c\right)_{n}}{\left(c-a-b\right)_{n}}},$$ whenever $\displaystyle c+d = a+b+1-n, \quad n=0,1,...$

Therefore $$\begin{align*} S = {}_{3}F_{2} \left[ {-n, -n, -n \atop -3n, 1}; 1 \right] =& \frac{(-2n)_{n}(-2n)_{n}}{(-3n)_{n}(-n)_{n}}\\ =& \frac{(-1)^n(n+1)_{n}(-1)^n(n+1)_{n}}{(-1)^n(2n+1)_{n}(-1)^n(1)_{n}} \quad \textrm{ from (*)} \\ =& \frac{(n+1)_{n}(n+1)_{n}}{(2n+1)_{n}(1)_{n}}\\ =& \frac{\Gamma(2n+1)\Gamma(2n+1)\Gamma(2n+1)}{\Gamma(n+1)\Gamma(n+1)\Gamma(3n+1)n!}\\ =& \frac{(2n)!^3}{n!^3(3n)!} \end{align*}$$ Therefore we can conclude $$\boxed{ \sum_{k=0}^{n} \frac{{\binom{n}{k}}^3}{\binom{3n}{k}} = \frac{(2n)!^3}{n!^3(3n)!}}$$