Finite series and product of Gammas

[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

 

[fresh_42]

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

 

[mfb]

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:

A046161(n)/a(n) = 1, 2/3, 8/15, 16/35, 128/315, 256/693, ... is binomial transform of Madhava-Gregory-Leibniz series for Pi/4 (i.e., 1 - 1/3 + 1/5 - 1/7 + ... ).

 

[ShayanJ]

Just for all who like to solve this riddle like me, the statement is equivalent to:

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

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

 

[mfb]

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

Can you give more terms?

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

Who is saying that?

 

[fresh_42]

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

No. The factor $2$ has been canceled by setting $d=2n$. (It's also missing in $2 \cdot 4 \cdot \ldots \cdot 2k$)

Also, can you give a reference for the RHS expressions?

I took the formulas from the German Wiki page of the Gamma function.

 

[ShayanJ]

Can you give more terms?

Never mind, he clarified.

Who is saying that?

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.

 

[mfb]

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.

 

[fresh_42]

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.