When I was trying to find closed-form representations for odd zeta-values, I used $$ \Gamma(z) = \frac{e^{-\gamma \cdot z}}{z} \prod_{n=1}^{\infty} \Big( 1 + \frac{z}{n} \Big)^{-1} e^{\frac{z}{n}} $$ and rearranged it to $$ \frac{\Gamma(z)}{e^{-\gamma \cdot z}} = \prod_{n=1}^{\infty} \Big( 1 + \frac{z}{n} \Big)^{-1} e^{z/n}. $$ As we know that $$\prod_{n=1}^{\infty} e^{z/n} = e^{z + z/2 + z/3 + \cdots + z/n} = e^{\zeta(1) z},$$ we can state that $$\prod_{n=1}^{\infty} \Big( 1 + \frac{z}{n} \Big) = \frac{e^{z(\zeta(1) - \gamma)}}{z\Gamma(z)}\qquad\text{(1)}$$ I then stumbled upon the Wikipedia page of Ramanujan Summation (see the bottom of the page), which I used to set $\zeta(1) = \gamma$ (which was, admittedly, a rather dangerous move. Remarkably, things went well eventually. Please don't stop reading). The $z^3$ -coefficient of both sides can now be obtained. Consider \begin{align*} (1-ax)(1-bx) &= 1 - (a+b)x + abx^2\\ &= 1-(a+b)x + (1/2)((a+b)^2-(a^2+b^2)) \end{align*} and \begin{align*}(1-ax)(1-bx)(1-cx) &= 1 - (a + b + c)x\\ &\qquad + (1/2)\Bigl((a + b + c)^2 - (a^2 + b^2 + c^2)\Bigr)x^2\\ &\qquad -(abc)x^3. \end{align*}

We can also set \begin{align*} (abc)x^3 &= (1/3)\Bigl((a^3 + b^3 + c^3) - (a + b + c)\Bigr)\\ &\qquad + (1/2)(a + b + c)^3 -(a + b + c)(a^2 + b^2 + c^2). \end{align*} It can be proved by induction that the x^3 term of $(1-ax)(1-bx)\cdots(1-nx)$ is equal to \begin{align*} (1/3)&\Bigl((a^3 + b^3 + c^3 +\cdots + n^3) - (a + b + c + \cdots + n^3))\\ &\qquad+ (1/2)(a + b + c + \cdots + n)^3 -(a + b + c + \cdots + n)(a^2 + b^2 + c^2 + \cdots + n^2).\qquad\text{(2)} \end{align*} On the right side of equation (1), the $z^3$-term can be found by looking at the z^3 term of the Taylor expansion series of $1/(z \Gamma(z))$, which is $(1/3)\zeta(3) + (1/2)\zeta(2) + (1/6)\gamma^3$. We then use (2) to obtain the equality $$ (1/6)\gamma^3 - (1/2)\gamma \pi^2 - (1/6) \psi^{(2)}(1) = 1/3)\zeta(3) + (1/2)\zeta(2) + (1/6)\gamma^3$$ to find that $$\zeta(3) = - (1/2) \psi^{(2)} (1),$$ (3) which is a true result that has been known (known should be a hyperlink but it isn't for some reason) for quite a long time. The important thing here is that I used $\zeta(1) = \gamma$, which isn't really true. Ramanujan assigned a summation value to the harmonic series (again, see Ramanujan Summation), and apparently it can be used to verify results and perhaps to prove other conjectures/solve problems.

My first question is: Is this a legitimate way to prove (3) ?

Generalizing this question:

When and how are divergent series and their summation values used in mathematics? What are the 'rules' when dealing with summed divergent series and using them to (try to) find new results?



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    $\begingroup$ I'm not sure why you think you can write the product formula for $\Gamma$ in terms of $\zeta(1)$ like you did, because $\zeta(1) = \infty$, so your formula doesn't make sense. I think you have to consider this only in the limiting case, and then make some argument that you are allowed to use ramanujan summation instead of normal summation, but still get that both sides are equal. $\endgroup$ – Eric Haengel Nov 19 '10 at 19:05
  • $\begingroup$ @Max: What is $\varphi$ $\endgroup$ – anonymous Nov 19 '10 at 19:06
  • $\begingroup$ @ Chandru1: $\varphi^{(n)}(x)$ is the n-th logarithmic deriviative of the gamma-function. Wikipedia has a good article on it. $\endgroup$ – Max Muller Nov 19 '10 at 19:10
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    $\begingroup$ @Eric Haengel: I would rather say $\zeta(1)$ is not defined - also limit values of $\zeta(s)$ as $s\to 1$ depends on how $s\to 1$. $\endgroup$ – AD. Nov 19 '10 at 23:00
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    $\begingroup$ You can try to contact Eric Delabaere, who's mentioned in the Wikipedia page. Also, please don't be discouraged with all the comments here! Sometimes these methods can be made rigorous, or alternatively can be converted to rigorous proofs by considering the meaning of the "summation convention". $\endgroup$ – Yuval Filmus Nov 20 '10 at 6:51

Since your last question looks much general: you might be interested in the nice introductory paper "Dances between discrete and continuous". Here D. Pengelley discusses some early motivations (and techniques how) to deal with divergent series (focused on L.Euler). I liked that paper much when I started reading about divergent summation.

Edit: Thinking a second time about your qualified question I can't imagine you don't know this all (basic reference G.H.Hardy, "divergent series" et al.) and you are looking for a more modern historical overview over that subject - so my answer might be somehow misfocused...

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  • $\begingroup$ @ Gottfried: thanks for the reference. I certainly didn't know all of it. A nice modern treatment of this subject that deals with my question would be much appreciated indeed. $\endgroup$ – Max Muller Nov 21 '10 at 11:35
  • $\begingroup$ This looks promising: books.google.nl/… $\endgroup$ – Max Muller Dec 1 '10 at 20:14

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