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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.)

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If you're looking for neat product formulae for $\zeta(2k+1)$, try investigating the infinite matrix product constructed by Bill Gosper. That gets you a bunch of values at once, depending on the size of the matrix. – J. M. Nov 2 '10 at 15:48
I have no idea how you got that last line from the preceding two. – Qiaochu Yuan Nov 2 '10 at 15:53
@ J.M. thank you for the suggestion, but I only have a very vague idea of what a matrix is... (I'm not enrolled in a university yet). Have people investigated the matrixes constructed by Gosper already? If so, I guess the values extracted from them have already been published. I'm looking for 'new' values. Have mathematicians sought to find new values by means of the method you described, already? – Max Muller Nov 2 '10 at 15:56
@ Qiaochu Yuan: when I check it now, me neither. I will correct it. – Max Muller Nov 2 '10 at 15:59
What do you mean, new values? New values of Riemann ζ at the odd integers? In any event, I suggest taking a look at Steven Finch's [ Mathematical Constants ](; you might find something interesting there. – J. M. Nov 2 '10 at 16:05
up vote 6 down vote accepted

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.

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@Qiaochu Yuan, you said an interesting thing "If anything simple worked, Euler would have done it, or somebody in the last few centuries anyway." -- isn't the proof of irrationality of $\zeta(3)$ contradicting this rule? – anon Nov 2 '10 at 16:17
@ Qiaochu Yuan: you're right. What a pitty :( . The other parts of the question still stand, though, although they may not have a definite answer. – Max Muller Nov 2 '10 at 16:19
@muad: Apery's proof of the irrationality of zeta(3) is not simple. – Qiaochu Yuan Nov 2 '10 at 16:20
@ Qiaochu: if we ignore having used the wrong infinite series, does it matter if the x and x^2 are present, to? Can we still equate the x^3 terms? – Max Muller Nov 2 '10 at 16:36
@Max Muller: the fact that the x and x^2 terms are wrong in what you've written indicates that the identity is wrong. Once you fix the identity, the x and x^2 terms will also match up, but again I expect that equating the x^3 terms won't tell you anything new. – Qiaochu Yuan Nov 2 '10 at 16:38

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