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This question was posed originally on MSE, I put it here because I didn't receive the answer(s) I wished to see.

Dear MO-Community,

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\Big)\\\ &\qquad+ (1/2)\Big((a + b + c + \cdots + n)^3 -(a + b + c + \cdots + n)(a^2 + b^2 + c^2 + \cdots + n^2)\Big).\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?

Thanks,

Max

As I suspect someone (I was thinking of Qiaochu Yuan himself) will add this too, I will add this question for him/her, as it is somewhat related.

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8  
+1 for "please don't stop reading". (I will introduce it in the middle of the next paper) –  Pietro Majer Nov 22 '10 at 23:07
    
Sorry, which is equation (2) ? would you add the eqation numbers? Also, what is $\psi^(2)$? –  Pietro Majer Nov 22 '10 at 23:13
    
@ Pietro Majer: Equation (2) is the x^3 term of (1-ax)(1-bx)...(1-nx). The (2) is hidden behind/next to the "related questions". –  Max Muller Nov 22 '10 at 23:13
    
For the second question, see the Polygamma function article on Wikipedia. –  Max Muller Nov 22 '10 at 23:15
1  
Max, do you have enough background to read a beginning text on complex analysis? Maybe somebody can suggest one that gets quickly to product and series developments. The basic example of value assignment for divergent series is not an assignment at all, it is simply reporting the value of analytic continuation of the function at a point outside the open disk of guaranteed convergence. Other meanings have sometimes occurred, but your idea of finding a divergent series or product and giving the most pleasing value is, well, not good. –  Will Jagy Nov 23 '10 at 6:46

3 Answers 3

RULES ... don't just use divergent series to get an answer unless there is some additional work to show it is meaningful. Indeed, don't even use rearrangement of conditionally convergent series and expect to get something meaningful (again) unless there is some additional work.

AN EXAMPLE ... Fourier series sometimes diverge, but converge when summed (C,1). There are theorems explaining conditions when the (C,1) sum of a Fourier series for a function actually converges to that function.

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Gerald Edgar: thanks for the answer. Do you have any ideas on how to show that $\zeta(1)=\gamma$, and therefore can be used in this context? –  Max Muller Nov 23 '10 at 16:04
    
Well, not to show that $\zeta(1)=\gamma$, but I am interested in the additional work you mentioned to make such equalities meaningful and useful? –  Max Muller Nov 23 '10 at 16:24
    
The best source is Hardy's book, DIVERGENT SERIES. The simplest of the "additional work" results are known as "Tauberian theorems". –  Gerald Edgar Nov 23 '10 at 17:03

It seems likely to me that your derivation can be modified to just involve convergent series and products, but I have not tried it.

There is however a well-known easy derivation of your identity $\zeta(3) = - (1/2) \psi^{(2)} (1)$ from the product formula, which is quite close to what you are doing. Note that $\psi=\Gamma'/\Gamma$. Take the logarithmic derivative of the product formula. Then the product becomes a series: $$\frac{\Gamma'}{\Gamma}(z)=-\gamma-\frac 1z+\sum_{n=1}^\infty\left(\frac 1n-\frac 1{z+n}\right)=-\gamma+\sum_{n=1}^\infty\left(\frac 1n-\frac 1{z+n-1}\right). $$ Now just differentiate twice and put $z=1$. All series and products involved converge nice enough so that termwise differentiation is justified.

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Without claiming to check your algebra, you appear to be engaging in zeta function regularization.

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@ Steve Huntsman: thanks for the answer, I haven't read all the info on the link yet but I will do so tomorrow. (It's getting late). –  Max Muller Nov 22 '10 at 22:57
    
the article mentions that regularized sums are used to obtain results for conditionally convergent sums, so I think we could count that as an application of summed divergent series... +1. –  Max Muller Nov 23 '10 at 16:28

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