# Two series involving the Gamma function

The last piece I am left with in my proof is to compute the following two series:

$$\sum_{i=1}^{n-1}\dfrac{\Gamma(i-d)\Gamma(n-i+d)}{\Gamma(i+1)\Gamma(n-i)(n-i-d)}, \sum_{i=1}^{n-1} \dfrac{\Gamma(n-d-i)\Gamma(i+d)}{(i-d)\Gamma(i+1)\Gamma(n-i)},$$

here $n\geq 2$, $d\in(-0.5, 0.5)$. According to Maple that would be

\begin{align} & \sum_{i=1}^{n-1}\dfrac{\Gamma(i-d)\Gamma(n-i+d)}{\Gamma(i+1)\Gamma(n-i)(n-i-d)}= \\ &=\dfrac{\Gamma(-d)}{2}\left[\dfrac{d\Gamma(-d)\Gamma(d)\Gamma(n-2d)\Gamma(n)-2\Gamma(n+d)\Gamma(n-d)\Gamma(-2d)}{(n-d)\Gamma(n)\Gamma(n-d)\Gamma(-2d)}\right] \\ &= \dfrac{d\Gamma(-d)^2\Gamma(d)\Gamma(n-2d)}{2\Gamma(n-d+1)\Gamma(-2d)}-\dfrac{\Gamma(-d)\Gamma(n+d)}{(n-d)\Gamma(n)} \end{align} and \begin{align} &\sum_{i=1}^{n-1} \dfrac{\Gamma(n-d-i)\Gamma(i+d)}{(i-d)\Gamma(i+1)\Gamma(n-i)}= \\ &=\dfrac{\Gamma(d)}{2}\dfrac{d\Gamma(-d)^2\Gamma(n-2d)\Gamma(n)+2\Gamma(n-d)^2\Gamma(-2d)}{d\Gamma(-2d)\Gamma(n-d)\Gamma(n)} \\ &= \dfrac{\Gamma(d)\Gamma(-d)^2\Gamma(n-2d)}{2\Gamma(-2d)\Gamma(n-d)} + \dfrac{\Gamma(n-d)\Gamma(d)}{d\Gamma(n)}. \end{align}

I have already simplified these series as much as possible and could not find any more elementary properties of the $\Gamma$ function that would make these expressions more convenient. Therefore, I thought I should use one of the definitions of the $\Gamma$ function. Also I believe it should be something that involves integrals or sums (rather than products) in order to change the order of summation and apply $\sum_{i=1}^{n-1}$ first.

However, writing the first series as $$\sum_{i=1}^{n-1}\dfrac{\int_0^\infty e^{-u}u^{i-d-1}du\int_0^\infty e^{-v}v^{n-i+d-1}dv}{(n-i-d)\int_0^\infty e^{-z}z^idz\int_0^\infty e^{-w}w^{n-i-1}dw\int_0^\infty e^{-r}r^{n-i-d-1}dr} =$$ $$\sum_{i=1}^{n-1}\dfrac{\int^\infty_0 \int^\infty_0 e^{-u-v}u^{i-d-1}v^{n-i+d-1}dudv}{(n-i-d) \int^\infty_0 \int^\infty_0 \int^\infty_0 e^{-z-w-r}z^{i}w^{n-i-1}r^{n-i-d-1}dzdwdr}$$

does not help at all because of denominator.

I also tried to rewrite it (both the series and the desired result) using the Beta function: \begin{align} \sum_{i=1}^{n-1}\dfrac{\Gamma(i-d)\Gamma(n-i+d)}{\Gamma(i+1)\Gamma(n-i)(n-i-d)} &= \sum_{i=1}^{n-1}\dfrac{B(d+1,i-d)B(n-i+d,1-d)B(n-i-d,1) (i-n)}{\Gamma(1+d)\Gamma(-d)d}\\ &= \dfrac{B(-d,-d)B(n-2d,d)B(n-i-d,1) + 2B(-d,n+d)}{2(n-d)} \end{align} and to use a couple of definitions: $$B(x,y) = 2\int_0^{\pi/2}(\sin\theta)^{2x-1}(\cos\theta)^{2y-1}\,d\theta, \qquad \mathrm{Re}(x)>0,\ \mathrm{Re}(y)>0,$$ $$B(x,y) = \int_0^\infty\dfrac{t^{x-1}}{(1+t)^{x+y}}\,dt, \qquad \mathrm{Re}(x)>0,\ \mathrm{Re}(y)>0.$$ However, even though they allowed me to use geometric progression, resulting expressions were too cumbersome (or should I try harder?). Even if this is a correct way, I cannot see how to go from a triple integral (geometric progression, cumbersome expression) to a triple + single integral, i.e. $$\dfrac{B(-d,-d)B(n-2d,d)B(n-i-d,1) + 2B(-d,n+d)}{2(n-d)}$$ (here I ignore requirements $\mathrm{Re}(x)>0$, $\mathrm{Re}(y)>0$, which can be easily satisfied).

In case all this is not terribly difficult I would like only a hint first.

This does not really help, but since there seems to be some particular way to evaluate this kind of series I tried to take some simpler one first and just got annoyed by such an elegant answer (the one I got from Maple, of course):

$$\sum_{i=1}^{n-1}\dfrac{\Gamma(i-d)\Gamma(n-i+d)i}{\Gamma(i+1)\Gamma(n-i)}=\Gamma(1-d)\Gamma(1+d)(n-1)=\dfrac{d(1-n)\pi}{\sin(\pi(1+d))}$$

(comparing with the very first series above $\frac{1}{n-d-i}$ here is just replaced with $i$).

Edit: the desired results also can be rewritten to

\begin{align} & \sum_{i=1}^{n-1}\dfrac{\Gamma(i-d)\Gamma(n-i+d)}{\Gamma(i+1)\Gamma(n-i)(n-i-d)} \\ &= \dfrac{\Gamma(-d)\Gamma(n+d)}{(n-d)\Gamma(n)} \left(\dfrac{d\Gamma(2d)\Gamma(n)\Gamma(d-n+1)}{\Gamma(n+d)\Gamma(2d-n+1)} - 1\right), \\ &\sum_{i=1}^{n-1} \dfrac{\Gamma(n-d-i)\Gamma(i+d)}{(i-d)\Gamma(i+1)\Gamma(n-i)} \\ &=-\dfrac{\Gamma(d)\Gamma(n-d)}{d\Gamma(n)} \left(\ \dfrac{\Gamma(n)\Gamma(n-2d)\Gamma(1-d)^2}{\Gamma(1-2d)\Gamma(n-d)^2} -1\right). \end{align}

• The numerator has $\Gamma(i-d)\Gamma(n-[i-d])$, which might perhaps be rephrased using a combination between the reflection formula and the multiplication formula. – Lucian Dec 24 '13 at 16:07
• @Lucian, thank you for an idea. I might be missing something, but I cannot see how to use the multiplication formula in this case. It seems that after applying $\Gamma(s+1)=s\Gamma(s)$ multiple times we are able to use the reflection formula, but then we are also left with a product $(n-i+d-1)\dots(1-i+d)$ which does not help since there is a single term in the denominator that contains $d$. – Julius Dec 24 '13 at 22:20

We can use the Generalized Hypergeometric function \begin{equation*} \,{}_pF_q(a_1,\ldots,a_p;b_1,\ldots,b_q;z) = \sum_{n=0}^\infty \frac{(a_1)_n\dots(a_p)_n}{(b_1)_n\dots(b_q)_n} \, \frac {z^n} {n!}, \end{equation*} with Pochhammer symbol defined by \begin{align*} (a)_0 &= 1, \\ (a)_n &= a(a+1)(a+2)...(a+n-1)=\dfrac{\Gamma(a+n)}{\Gamma(a)}, && n \geq 1, \end{align*} and evaluate the first series in the following way: \begin{align*} &\sum_{i=1}^{n-1}\dfrac{\Gamma(i-d)\Gamma(n-i+d)}{\Gamma(i+1)\Gamma(n-i)(n-i-d)} \\ &= \dfrac{\Gamma(1-d)\Gamma(n-1+d)}{\Gamma(n-1)(n-1-d)} {}_4F_3(1, 1 - d, 2 - n, 1 + d - n; 2, 2 + d - n, 2 - d - n; 1)) \\ &= \dfrac{\Gamma(1-d)\Gamma(n-1+d)}{\Gamma(n-1)(n-1-d)} \sum_{m=0}^\infty \dfrac{(1)_m(1-d)_m(2-n)_m(1+d-n)_m}{(2)_m(2+d-n)_m(2-d-n)_m}\dfrac{1}{m!} \\ &= \dfrac{\Gamma(1-d)\Gamma(n-1+d)}{\Gamma(n-1)(n-1-d)} \sum_{m=0}^\infty \dfrac{(1-d)_m(2-n)_m(1+d-n)_m}{(m+1)(2+d-n)_m(2-d-n)_m}\dfrac{1}{m!} \\ &= \dfrac{\Gamma(1-d)\Gamma(n-1+d)}{\Gamma(n-1)(n-1-d)} \sum_{m=0}^\infty \dfrac{(1-d)_m(2-n)_m(1+d-n)_m}{(2+d-n)_m(2-d-n)_m}\dfrac{1}{(m+1)!} \\ &= \dfrac{\Gamma(1-d)\Gamma(n-1+d)}{\Gamma(n-1)(n-1-d)} \dfrac{(1+d-n)(1-n-d)}{-d(1-n)(d-n)}\sum_{m=0}^\infty \dfrac{(-d)_{m+1}(1-n)_{m+1}(d-n)_{m+1}}{(1+d-n)_{m+1}(1-d-n)_{m+1}}\dfrac{1}{(m+1)!} \\ &= \dfrac{\Gamma(-d)\Gamma(n+d)}{(n-d)\Gamma(n)} ({}_3F_2(-d,1-n,d-n;-d-n+1,d-n+1;1)-1) \\ &= \dfrac{\Gamma(-d)\Gamma(n+d)}{(n-d)\Gamma(n)} \left(\dfrac{(2d-n+1)_{n-1}(1)_{n-1}}{(d-n+1)_{n-1}(d+1)_{n-1}}-1\right) \\ &= \dfrac{\Gamma(-d)\Gamma(n+d)}{(n-d)\Gamma(n)} \left(\dfrac{\Gamma(2d)\Gamma(n)\Gamma(d-n+1)\Gamma(d+1)}{\Gamma(d)\Gamma(n+d)\Gamma(2d-n+1)} - 1\right)\\ &= \dfrac{\Gamma(-d)\Gamma(n+d)}{(n-d)\Gamma(n)} \left(\dfrac{d\Gamma(2d)\Gamma(n)\Gamma(d-n+1)}{\Gamma(n+d)\Gamma(2d-n+1)} - 1\right),\\ \end{align*} here we used Saalschütz's theorem: \begin{equation*} {}_3F_2 (a,b, -n;c, 1+a+b-c-n;1)= \frac{(c-a)_n(c-b)_n}{(c)_n(c-a-b)_n}. \end{equation*}

In the same way we evaluate the second series: \begin{align*} &\sum_{i=1}^{n-1} \dfrac{\Gamma(n-d-i)\Gamma(i+d)}{(i-d)\Gamma(i+1)\Gamma(n-i)} \\ &= \dfrac{\Gamma(1+d)\Gamma(n-d-1)}{\Gamma(n-1)(1-d)} {}_4F_3(1, 1-d, 2-n, 1+d; 2, 2-d, d-n+2; 1)) \\ &= \dfrac{\Gamma(1+d)\Gamma(n-d-1)}{\Gamma(n-1)(1-d)} \dfrac{(1-d)(d-n+1)}{d^2(n-1)}({}_3F_2(-d,d,1-n;1-d,d-n+1,1)-1) \\ &= -\dfrac{\Gamma(d)\Gamma(n-d)}{d\Gamma(n)} \left(\ \dfrac{(1)_{n-1}(1-2d)_{n-1}}{((1-d)_{n-1})^2} -1\right) \\ &= -\dfrac{\Gamma(d)\Gamma(n-d)}{d\Gamma(n)} \left(\ \dfrac{\Gamma(n)\Gamma(n-2d)\Gamma(1-d)^2}{\Gamma(1-2d)\Gamma(n-d)^2} -1\right). \\ \end{align*}

I'm not sure how useful this hint will be, but here it is:

$$\text{Since we know that}\;\;\Gamma(s+1)=s\Gamma (s)\;,\;\;\text{we get that}$$

$$\dfrac{\Gamma(i-d)\Gamma(n-i+d)}{\Gamma(i+1)\Gamma(n-i)(n-i-d)}=\frac{\color{red}{\Gamma(i-d)}\cdot\prod_{k=1}^d(n-i+d-k)\color{green}{\Gamma(n-i)}}{\prod_{k=0}^d(k-d)\cdot\color{red}{\Gamma(i-d)}\color{green}{\Gamma(n-i)}\Gamma(n-i-d)}$$

• Just added to my question that $d\in(-0.5, 0.5)$ so, unfortunately, this does not apply. – Julius Dec 24 '13 at 16:07
• Well, it still applies but it doesn't come out as pretty as expected...:) – DonAntonio Dec 24 '13 at 16:09