# A Finite series and product of Gammas

1. Apr 28, 2017

### ShayanJ

In reviewing some calculations, I've arrived at the series:

$S(d)=-\frac 1 {d-1}+\frac 1 2 \frac{d-2}{d-3}-\frac 1 8 \frac{(d-2)(d-4)}{d-5}+\frac 1 {48} \frac{(d-2)(d-4)(d-6)}{d-7}+\dots$

Its an infinite series but because I'm interested in its values for even $d$s, its actually a finite series at each evaluation. Its values are:

$\begin{array}{c|ccccc} d \ \ \ &2 \ \ \ &4 \ \ \ &6 \ \ \ &8 \ \ \ &10 \ \ \ &12 \\\hline S(d) \ \ \ &-1 \ \ \ &\frac 2 3 \ \ \ &-\frac 8 {15} \ \ \ &\frac{16}{35} \ \ \ &-\frac{128}{315} \ \ \ &\frac{256}{693}\end{array}$

But the interesting fact is that the expression $G(d)=\frac{\Gamma(\frac d 2)\Gamma(\frac{1-d}2)}{2 \sqrt \pi}$ has the exact same values for at least the above values of $d$. So it seems that I should be able to prove that $S(d)=G(d)$. But I have no idea how to do it. At first I tried to somehow write the series in terms of the Hypergeometric function, but that is both impossible and useless. So, any ideas?

EDIT: Another evidence is that both $S(d)$ and $G(d)$ are singular for odd $d$s.

Thanks

Last edited: Apr 28, 2017
2. Apr 28, 2017

### Staff: Mentor

Just for all who like to solve this riddle like me, the statement is equivalent to:
$$- \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!} \cdot \frac{1}{2n-(2k+1)}\cdot \prod_{i=1}^{k}(n-i) = \frac{n!(n-1)!}{2(2n)!}\cdot (-4)^n = (-1)^n \cdot \frac{2^{2n}}{(2n)\binom{2n}{n}}$$

Last edited: Apr 28, 2017
3. Apr 28, 2017

### Staff: Mentor

The denominators also appear in the series expansion of $(1-x)^{-3/2}$.

OEIS sequence. We even get the numerators there with this interesting comment:

4. Apr 28, 2017

### ShayanJ

It should be ...(n-2i)...

Anyway, are you guys saying that this is a famous unsolved problem?

5. Apr 28, 2017

### Staff: Mentor

Can you give more terms?
Who is saying that?

6. Apr 28, 2017

### Staff: Mentor

No. The factor $2$ has been canceled by setting $d=2n$. (It's also missing in $2 \cdot 4 \cdot \ldots \cdot 2k$)
I took the formulas from the German Wiki page of the Gamma function.

7. Apr 28, 2017

### ShayanJ

Never mind, he clarified.

I just got that impression, specially because of your post since I'm not quite sure how you could find that information without previous knowledge of the problem.

8. Apr 28, 2017

### Staff: Mentor

I searched for the first denominators on OEIS, then calculated more using fresh's formula to find the right sequence, and confirmed it via the given formulas there.

9. Apr 29, 2017

### Staff: Mentor

I don't know, whether it's already obvious to you, but in any case and for the records: with an induction argument (or simply a recursion on the right hand side), it boils down to show
$$- \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!} \cdot \frac{1}{d-(2k+1)}\cdot \prod_{i=1}^{k}(\frac{d}{2}-i) = (-1)^{\frac{d}{2}} \cdot \frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} \cdot \ldots \cdot \frac{d-2}{d-1}$$
which itself can very likely be shown by another induction.