Why is the following evaluation of Apery's Constant wrong and do you have suggestions on how, if at all, this method could be improved?

Question

Please let me summarize the method by which L. Euler solved the Basel Problem and how he found the exact value of $\zeta(2n)$ up to $n=13$. Euler used the infinite product
$$ \displaystyle f(x) = \frac{\sin(x)}{x} = \prod_{n=1}^{\infty} \Big(1-\frac{x^2}{n^2\pi^2}\Big) , $$ Newton's identities and the (Taylor) Series Expansion (at $x=0$) of the sine function divided by $x$ to arrive at $$ 1 - \frac{x^2}{\pi^2} \cdot (1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + ... + \frac{1}{n^2}) + x^4(...) = 1 - \frac{x^2}{6} + \frac{x^4}{120} - ... $$ Upon subtracting 'one' from both sides, equating the $x^2$ terms to each other and multiplying both sides by $ - \pi^2$, one finds that $$ \zeta(2)=\frac{\pi^2}{6}. $$ When I first saw this proof and the way it was extended to find the values of the other even zeta-constants, I couldn't help myself thinking: "How could this method be strenghtened to find the values of the odd zeta-constants?" (And, a little while later, "why hasn't this been done before?")

I started looking for a similar-looking infinite product, only now I focussed on one of the form $$ \displaystyle f(x) = \prod_{n=1}^{\infty} \Big(1-\frac{x^a}{k^3 \cdot q}\Big) $$ (for some $ a \in \mathbb{N} , q \in \mathbb{R} $). A little while later I stumbled upon this website and fixated my eyeballs on equation (27). If we take $n=3$, Prudnikov et al. tell us that $$ \prod_{n=1}^{\infty} \Big(1-\frac{x^3}{k^3}\Big) = - \frac{1}{x^3} \cdot \prod_{k=1}^{2} \frac{1}{\Gamma(-e^{2/3 \pi i k} \cdot x)}. $$ Now, I thought that if we could use Newton's Identities again on the left side of the equation and find out what the Taylor Series Expansion of the right-hand side would be, we could find out what the exact value of Apery's Constant and other odd zeta-constants would be. In this answer by Robert Smith, I was told the Series Expansion. So we have $$ 1 - x^3(1 + \frac{1}{8} + \frac{1}{27} + ... + \frac{1}{n^3}) = -1 - 2 \cdot \gamma x - 2 \gamma^2 x^2 + \frac{1}{6}x^3(-8\gamma^3 - \psi^{(2)} (1)) - x^4(...) $$ Notice that on the left side we only have 'one minus a term with an $x^3$ coefficient', while on the other side we see 'minus one plus $x$, $x^2$, $x^3$ coefficients with their terms'. This is important, because it probably answers the question why the following will not work, but I don't know why and I really would like to know.

I guess you know what I will attempt to do now. We equate the $x^3$ terms with each other, set $x=1$, multiply by minus one and 'find' that $$ \zeta(3) = \frac{1}{6}(8\gamma^3 + \psi^{(2)} (1)). $$ By combining this with the already known result $$ \zeta(3) = -\frac{1}{2} \psi^{2}(1), $$ we 'find' that $$ \zeta(3) '=' \gamma^3. $$ Obviously, this is wrong. Apery's constant is larger than one, and this value is clearly smaller than one. Could someone please elaborate one where I went wrong? And does anybody have any sugguestions and/or ideas related to the discussion from above using which we could find "better" values for Apery's Constant and the other odd zeta constants? (For example by pointing out a similar infinite product relation, and by showing that that infinite product has a nicer Series Expansion?) Or could someone point out to me why this approach to finding nicer closed-form representations for these constants clearly won't lead to any results?

Thanks in advance,

Max Muller

(Moderators: If you find any spelling mistakes or errors grammar errors, feel free to correct them. To the rest: $\gamma$ is the Euler-Mascheroni Constant, and it amounts to approximately $0.5772$. The $\psi^{(2)}(x)$ stands for the second logarithmic deriviative of the Gamma-function. As usual, Wikipedia is a pretty good reference for this sort of things.)

Answer 1

Your previous question was about the wrong function. Instead of $\Gamma(x)$ in the denominator you should have $\Gamma(-x)$. If you fix this you'll probably end up with the same polygamma identity you already knew.

In any case, proving anything about $\zeta(2k+1)$ is known to be quite hard. If anything simple worked, Euler would have done it, or somebody in the last few centuries anyway.

Let me also mention that Prudnikov's identity is, if you are willing to accept Euler-style manipulations, trivial. It is equivalent to a product formula for $\frac{1}{\Gamma(x)}$ which follows (again if you are willing to accept Euler-style manipulations) from an investigation of its roots and does not really tell you anything deep about zeta values.