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I was looking at this identities in Wikipedia

$$\frac{\zeta(3 / 2)}{\zeta(-1 / 2)}=-4 \pi$$ $$\frac{\zeta(5 / 2)}{\zeta(-3 / 2)}=-\frac{16 \pi^2}{3} $$ $$\frac{\zeta(7 / 2)}{\zeta(-5 / 2)}=\frac{64 \pi^3}{15} $$ $$\frac{\zeta(9 / 2)}{\zeta(-7 / 2)}=\frac{256 \pi^4}{105}$$

And wondering if if there is a general term.

Is there a general term? If there is, what its is and how to obtain it.

Neves
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2 Answers2

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If you use the functional equation of the zeta function, it satisfies $$ \zeta(s)=2^s\pi^{s-1}\sin(\frac{\pi s}{2})\Gamma(1-s)\zeta(1-s) $$ Now observe that if $s=\frac{2n+1}{2}$, then $1-s=1-\frac{2n+1}{2}=\frac{1-2n}{2}$. Thus, the functional equation implies that $$ \frac{\zeta\left(\frac{2n+1}{2}\right)}{\zeta\left(\frac{1-2n}{2}\right)}=2^{\frac{2n+1}{2}}\pi^{\frac{2n-1}{2}}\sin\left(\frac{\pi(2n+1)}{4}\right)\Gamma\left(\frac{1-2n}{2}\right) $$

Steven Creech
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  • If it matters at all, thanks to the constraints on the problem / the type of $s$ inputted, one can generate a more specific answer without a dependence on sine or the gamma function (my answer explains this). – PrincessEev Aug 02 '23 at 00:10
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Note: Errors existed in the previous version of this post. I'm including this notification to illuminate that the errors were fixed.

Well, the functional equation immediately tells us that $$ \frac{\zeta(s)}{\zeta(1-s)} = 2^s \pi^{s-1} \sin \left( \frac{\pi s}{2} \right) \Gamma(1-s) $$ Ultimately, the key to these for $s=\frac{2n+1}{2}$ lies in the value of $$ \Gamma(1-s) = \Gamma \left( 1-\frac{2n+1}{2} \right) $$ One might wonder why I chose this expression instead of that of your denominator. The gamma function has a reflection formula (see Wikipedia), $$ \Gamma(s) \Gamma(1-s) = \frac{\pi }{\sin(\pi s)} \text{ for non-integer } s $$ So we then have $$ \Gamma \left( 1-\frac{2n+1}{2} \right) = \frac{\pi }{\sin \left( \pi \frac{2n+1}{2} \right) \Gamma \left( \frac{2n+1}{2} \right)} $$ This sine term is $(-1)^n$. It is known that $$ \Gamma \left( \frac{2n+1}{2} \right) = \Gamma \left( n + \frac 1 2 \right) = \frac{(2n-1)!!}{2^n} \sqrt \pi $$ (see Wikipedia).

Therefore, $$ \Gamma \left( 1-\frac{2n+1}{2} \right) = \frac{(-1)^n \pi }{\frac{(2n-1)!!}{2^n} \sqrt \pi} = \frac{(-1)^n 2^n \sqrt \pi }{ (2n-1)!! } $$ Hence, using the functional equation and simplifying,

$$\begin{align*} \frac{\zeta\left(\frac{2n+1}{2}\right)}{\zeta\left(1-\frac{2n+1}{2}\right)} &= 2^{\frac{2n+1}{2}} \pi^{\frac{2n+1}{2}-1} \sin \left( \frac{\pi}{2} \frac{2n+1}{2} \right) \frac{(-1)^n 2^n \sqrt \pi }{ (2n-1)!! } \\ &= 2^{\frac{4n+1}{2}} \pi^{n} \sin \left( \frac{2n+1}{4} \pi \right) \frac{(-1)^n }{ (2n-1)!! } \\ \end{align*}$$

One more simplification of sorts can be made: just by looking at what numbers it outputs, $$ \sin \left( \frac{2n+1}{4} \pi \right) = \frac{(-1)^{\lfloor n/2 \rfloor} }{\sqrt 2} $$ so using this we get

$$ \frac{\zeta\left(\frac{2n+1}{2}\right)}{\zeta\left(1-\frac{2n+1}{2}\right)} = \frac{(-1)^{n+\lfloor n/2 \rfloor} (4\pi)^n }{ (2n-1)!! } $$

Not that I can justify this one more than "by playing around", but we can see that $$ (-1)^{n+\lfloor n/2 \rfloor} = (-1)^{\lceil n/2 \rceil} $$ to simplify just a tad more:

$$ \frac{\zeta\left(\frac{2n+1}{2}\right)}{\zeta\left(1-\frac{2n+1}{2}\right)} = \frac{(-1)^{\lceil n/2 \rceil} (4\pi)^n }{ (2n-1)!! } $$

PrincessEev
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  • There is something wrong. Note that for $n=2$ the LHS is $-52.63789013914$ and the RHS is $-131.5947253479$. – Neves Aug 01 '23 at 23:23
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    Yes, I'm aware, hence the note at the start of the post. I can't seem to track down the error though. – PrincessEev Aug 01 '23 at 23:27
  • In the development of $\Gamma\left(1-\frac{2n+1}{2}\right)$ the $2^{n-1}$ should be $2^n$ and the term $2n+1$ where did it come from? – Neves Aug 01 '23 at 23:41
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    I just came to the same conclusion myself - that I misremembered the reflection formula and didn't even think to look at it when I cited it. I'll double check the answer I just got and edit accordingly if it works. – PrincessEev Aug 01 '23 at 23:43
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    @Neves -- Okay I've found the errors and fixed them. (And as a sanity check, I did indeed check the values listed in the OP/Wikipedia just to be sure and they match.) – PrincessEev Aug 02 '23 at 00:01