Question

`\displaystyle \rm \sum_(n = 1)^ \infty \sum_(m = 1)^( \infty ) ( \Gamma(n) \Gamma(m) \Gamma(x))/( \Gamma(n + m + x))`​

Answer 1

Recall the beta function definition and gamma identity,

`\displaystyle \mathrm{B}(x,y) = \int_0^1 t^(x-1) (1-t)^(y-1) \, dt = (\Gamma(x) \Gamma(y))/(\Gamma(x+y))`

Consider the sum

`\displaystyle S(x) = \sum_(m=1)^\infty (\Gamma(m) \Gamma(x))/(\Gamma(m+x))`

Compute it by converting the gammas to the beta integral, interchanging summation with integration, and using the sum of a geometric series.

`\displaystyle S(x) = \sum_(m=1)^\infty \mathrm{B}(m,x) \\\\ ~~~~ = \sum_(m=1)^\infty \int_0^1 t^(m-1) (1-t)^(x-1) \, dt \\\\ \int_0^1 (1-t)^(x-1) \sum_(m=1)^\infty t^(m-1) \, dt \\\\ ~~~~ = \int_0^1 (1-t)^(x-2) \, dt \\\\ ~~~~ = \int_0^1 t^(x-2) \, dt \\\\ ~~~~ = \frac1{x-1} = ((x-2)!)/((x-1)!) = (\Gamma(x-1))/(\Gamma(x))`

It follows that

`\displaystyle S(n+x) = \sum_(m=1)^\infty (\Gamma(m) \Gamma(n+x))/(\Gamma(m+n+x)) = (\Gamma(n+x-1))/(\Gamma(n+x))`

Now we compute the sum of interest. It's just a matter of introducing appropriate gamma factors to condense the double series into a single hypergeometric one.

Recall the definition of the generalized hypergeometric function,

`\displaystyle {}_pF_q \left(\left.\begin{array}{c} a_1,a_2,\ldots,a_p\\b_1,b_2,\ldots,b_q\end{array}\right\vert z\right) = \sum_(n=0)^\infty ((a_1)_n (a_2)_n \cdots (a_p)_n)/((b_1)_n (b_2)_2 \cdots (b_q)_n) (z^n)/(n!)`

where
`(a)_n` denotes the Pochhammer symbol, defined by

`\begin{cases}(0)_n = 1 \\ (a)_n = a(a+1)(a+2)\cdots(a+n-1) = (\Gamma(a+n))/(\Gamma(a))\end{cases}`

We'll be needing the following identities later.

`(1)_n = n! = (\Gamma(n+1))/(\Gamma(1))`

`(x)_n = (\Gamma(n+x))/(\Gamma(x))`

`(x+1)_n = (\Gamma(n+x+1))/(\Gamma(x+1))`

The
`m`-sum is

`\displaystyle \sum_(m=1)^\infty (\Gamma(m))/(\Gamma(n+m+x)) = \frac1{\Gamma(n+x)} \sum_(m=1)^\infty (\Gamma(m)\Gamma(n+x))/(\Gamma(n+m+x)) \\\\ ~~~~~~~~ = (S(n+x))/(\Gamma(n+x)) \\\\ ~~~~~~~~ = (\Gamma(n+x-1))/(\Gamma(n+x)^2)`

Then the double sum reduces to

`\displaystyle \sum_(n=1)^\infty \sum_(m=1)^\infty (\Gamma(n)\Gamma(m)\Gamma(x))/(\Gamma(n+m+x)) = \sum_(n=1)^\infty (\Gamma(n)\Gamma(x)\Gamma(n+x-1))/(\Gamma(n+x)^2)`

Rewrite the summand. We use the property
`\Gamma(x+1)=x\Gamma(x)` to convert to Pochhammer symbols.

`\displaystyle (\Gamma(n)\Gamma(x)\Gamma(n+x-1))/(\Gamma(n+x)^2) = (\Gamma(n)\Gamma(x)^2\Gamma(n+x-1))/(\Gamma(n+x)^2 \Gamma(x))`

`\displaystyle . ~~~~~~~~ = \frac1{x^2} (\Gamma(n)\Gamma(x+1)^2\Gamma(n+x-1))/(\Gamma(n+x)^2\Gamma(x))`

`\displaystyle . ~~~~~~~~ = \frac1{x^2} (\Gamma(n) (\Gamma(n+x))/(\Gamma(x)))/((\Gamma(n+x-1)^2)/(\Gamma(x+1)^2))`

`\displaystyle . ~~~~~~~~ = \frac1{x^2} ((n-1)! (x)_(n-1))/(\left[(1+x)_(n-1)\right]^2)`

Now in the sum, shift the index to start at 0, and introduce an additional factor of
`n!` to get the hypergeometric form.

`\displaystyle \sum_(n=1)^\infty \frac1{x^2} ((n-1)! (x)_(n-1))/(\left[(1+x)_(n-1)\right]^2) = \frac1{x^2} \sum_(n=0)^\infty ((n!)^2 (x)_n)/(\left[(1+x)_n\right]^2) \frac1{n!} \\\\ ~~~~~~~~ = \frac1{x^2} \sum_(n=0)^\infty ([(1)_n]^2 (x)_n)/(\left[(1+x)_n\right]^2) \frac1{n!} \\\\ ~~~~~~~~ = \boxed{\frac1{x^2} \, {}_3F_2\left(\left.\begin{array}{c}1,1,x\\1+x,1+x\end{array}\right\vert1\right)}`