# Challenge 1: Multiple Zeta Values

1. Sep 26, 2013

### Office_Shredder

Staff Emeritus
A multiple zeta value is defined as
$$\zeta(s_1,...,s_k) = \sum_{n_1 > n_2 ... > n_k > 0} \frac{1}{n_1^{s_1} n_2^{s_2}...n_k^{s_k}}$$.
For example,
$$\zeta(4) = \sum_{n = 1}^{\infty} \frac{1}{n^4}$$
and
$$\zeta(2,2) = \sum_{m =1}^{\infty} \sum_{n = 1}^{m-1} \frac{1}{ m^2 n^2}$$.

Prove the following relationship:
$$\zeta(2)^2 = 4 \zeta(3,1) + 2 \zeta(2,2)$$

2. Sep 26, 2013

### jackmell

This looks like when m=1, then n goes from 1 to 0 causing the denominator to be zero. Am I mis-interpreting your double sum and if so could you clarify please?

3. Sep 26, 2013

### Office_Shredder

Staff Emeritus
The intention is that there are no entries in the sum when m = 1 (if you check the general definition the smallest value n1 can be is 2 if k=2)

4. Sep 26, 2013

### jbunniii

I hope I'm not violating the rules by posting a partial solution. The FAQ is silent on the subject.

First we note that, by symmetry of $m$ and $n$,
$$\sum_{m=1}^{\infty}\sum_{n > m} \frac{1}{m^2 n^2} = \sum_{m=1}^{\infty}\sum_{n < m} = \frac{1}{m^2 n^2} = \zeta(2,2)$$
Therefore, we may write
\begin{align} \zeta(2)^2 &= \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m^2 n^2} \\ &= \sum_{m=1}^{\infty} \sum_{n<m} \frac{1}{m^2 n^2} + \sum_{m=1}^{\infty} \sum_{n>m} \frac{1}{m^2 n^2} + \sum_{n=1}^{\infty} \frac{1}{n^4} \\ &= 2\zeta(2,2) + \zeta(4)\end{align}
So the problem reduces to showing that $\zeta(4) = 4\zeta(3,1)$. I have a proof for this but the margin is too small to contain it. I'll see if I can find a shorter one.

5. Sep 27, 2013

### CompuChip

Be careful. Last time I heard someone say that it took about 3 centuries before they got it right.

6. Sep 27, 2013

### jbunniii

I've been playing around with the usual trick of expressing the series in terms of a power series so calculus techniques can be applied. Formally (without justifying the interchanges of sum and integral), we have:
\begin{align} \zeta(4) &= \sum_{n=1}^{\infty}\frac{1}{n^4}\\ &= \left. \sum_{n=1}^{\infty}\frac{x^n}{n^4}\right|_{x=1} \\ &= \sum_{n=1}^{\infty} \frac{1}{n^3} \left.\left(\frac{x^n}{n}\right)\right|_{x=1} \\ &= \sum_{n=1}^{\infty} \frac{1}{n^3} \int_{0}^{1} x^{n-1} dx \\ &= \int_{0}^{1} \sum_{n=1}^{\infty} \frac{x^{n-1}}{n^3} dx\\ &= \int_{0}^{1} \frac{1}{x} \sum_{n=1}^{\infty} \frac{x^n}{n^3} dx \end{align}
Now we can use almost the same trick with $\sum x^n / n^3$ as we did with $\sum x^n / n^4$. (Unfortunately we'll have a couple more iterations of this.)
\begin{align} &= \int_{0}^{1} \frac{1}{x} \sum_{n=1}^{\infty} \frac{1}{n^2} \left(\frac{x^n}{n}\right) dx\\ &= \int_{0}^{1} \frac{1}{x} \sum_{n=1}^{\infty} \frac{1}{n^2} \left(\int_{0}^{x} y^{n-1}dy\right) dx\\ &= \int_{0}^{1} \frac{1}{x} \int_{0}^{x} \sum_{n=1}^{\infty} \frac{y^{n-1}}{n^2} dy dx \\ &= \int_{0}^{1} \frac{1}{x} \int_{0}^{x} \frac{1}{y} \sum_{n=1}^{\infty} \frac{y^n}{n^2} dy dx \\ \end{align}
And here we go again...
\begin{align} &= \int_{0}^{1} \frac{1}{x} \int_{0}^{x} \frac{1}{y} \sum_{n=1}^{\infty} \frac{1}{n} \left(\frac{y^n}{n}\right) dy dx\\ &= \int_{0}^{1} \frac{1}{x} \int_{0}^{x} \frac{1}{y} \sum_{n=1}^{\infty} \frac{1}{n} \left(\int_{0}^{y}z^{n-1}dz\right) dy dx \\ &= \int_{0}^{1} \frac{1}{x} \int_{0}^{x} \frac{1}{y} \int_{0}^{y} \sum_{n=1}^{\infty} \frac{z^{n-1}}{n} dz dy dx \\ &= \int_{0}^{1} \frac{1}{x} \int_{0}^{x} \frac{1}{y} \int_{0}^{y} \frac{1}{z} \sum_{n=1}^{\infty} \frac{z^n}{n} dz dy dx \\ \end{align}
And one last time...
\begin{align} &= \int_{0}^{1} \frac{1}{x} \int_{0}^{x} \frac{1}{y} \int_{0}^{y} \frac{1}{z} \sum_{n=1}^{\infty} \int_{0}^{z} w^{n-1} dw dz dy dx \\ &= \int_{0}^{1} \frac{1}{x} \int_{0}^{x} \frac{1}{y} \int_{0}^{y} \frac{1}{z} \int_{0}^{z} \sum_{n=1}^{\infty} w^{n-1} dw dz dy dx\\ \end{align}
Now $\sum_{n=1}^{\infty}w^{n-1} = 1/(1-w)$ for $|w| < 1$, so the above reduces to
\begin{align} \int_{0}^{1} \frac{1}{x} \int_{0}^{x} \frac{1}{y} \int_{0}^{y} \frac{1}{z} \int_{0}^{z} \frac{1}{1-w} dw dz dy dx\\ \end{align}
Unfortunately after all this work, it's still not clear whether it will get us anywhere. But the next step would be to start manipulating $\zeta(3,1)$ in a similar way and see if we can get a match. Even then, we won't be done, as all the integral-sum interchanges will have to be justified. I think this will be no big deal as we can invoke the uniform convergence of power series.

7. Sep 29, 2013

### Office_Shredder

Staff Emeritus
That's a really nice calculation jbunniii - I thought I'd just give you a bit of encouragement by saying that you can get similar formulas for the other zeta values.

8. Sep 29, 2013

### jbunniii

I've spent some time obtaining a similar expression for $\zeta(3,1)$ but I haven't worked out how to manipulate it to the expression for $\zeta(4)$ with an extra factor of 4. I got tired of that after a while and tried a different approach using summation by parts, but that hasn't worked out yet either. Will keep playing with it as time allows.

9. Oct 1, 2013

### Office_Shredder

Staff Emeritus
jbunniii, the relation $\zeta(4) = 4\zeta(3,1)$ might be a red herring....

10. Oct 1, 2013

### jbunniii

11. Oct 1, 2013

### Mandelbroth

I don't think so. I'll start from the beginning.

I remember, from a kick I had in complex analysis, the beautiful formula $$\zeta(m,n)+\zeta(n,m)=\zeta(m)\zeta(n)-\zeta(m+n).$$

Letting $n=m=2$ gives $2\zeta(2,2)=\zeta(2)^2-\zeta(4)$.

After about 2 hours of getting nowhere with this, I broke down and asked the internet what it knew about multiple zeta functions. I found this source as a reference. I have used two of its theorems in the following.

To start, I use the Sum Theorem to obtain $\zeta(4)=\zeta(3,1)+\zeta(2,2)$. Then, I use the pretty little identity shown below the Derivation Theorem to show $3\zeta(3,1)=\zeta(2,2)$. Plugging this into the above formula, we get $\zeta(4)=\zeta(3,1)+3\zeta(3,1)=4\zeta(3,1)$.

Thus, $\zeta(2)^2=2\zeta(2,2)+4\zeta(3,1)$. Halmos.

12. Oct 1, 2013

### jbunniii

Mandelbroth - It seems like you're bringing in some pretty specialized machinery for this. Are you prepared to prove your Sum Theorem and your pretty little identity?

Also, I proved the $\zeta(m,n)+\zeta(n,m)=\zeta(m)\zeta(n)-\zeta(m+n)$ identity for the special case $m=n=2$ in post #4. No complex analysis required.

I've assumed thus far that there must be a relatively elementary proof that doesn't require developing an extensive amount of theory about the zeta function. Slicing and dicing the sums the right way, most likely. Office_Shredder's most recent hint suggests that the way I sliced it in the first post might not be the right way, so I'm going back to the beginning to try again.

13. Oct 1, 2013

### jbunniii

Oops, deleted my last message due to some indexing errors. I'll save it and post an edited version later.

14. Oct 1, 2013

### Mandelbroth

There are proofs in the references given in the link. :tongue:

I agree that there is, probably, a more elementary method. I'll keep working on it after I finish my Spanish homework, and I'll see where we are in the morning.

15. Oct 1, 2013

### Office_Shredder

Staff Emeritus
The intent of the question was that it was supposed to be self-contained as far as necessary knowledge of multiple zeta values (and there are multiple ways to solve the problem like this that I know of, have no fear!)

I will admit that was a solid combination of googling and applying of theorems.

Last edited: Oct 1, 2013
16. Oct 9, 2013

### Office_Shredder

Staff Emeritus
Well, since there hasn't been any posts here in a while I thought I would give a small prod by noting that another way of writing, for example, $\zeta(2,2)$ is
$$\zeta(2,2) = \sum_{m,n=1}^{\infty} \frac{1}{m^2(m+n)^2}$$
and there are similar expressions for the other multiple zeta values that should be obvious. Maybe this form is a bit easier to work with for what you guys are trying.

17. Nov 17, 2013

### Office_Shredder

Staff Emeritus
There still exists a completely elementary three line proof waiting to be found!

18. May 28, 2014

### jostpuur

$$\zeta(2,2) = \sum_{m=2}^{\infty} \sum_{n=1}^{m-1} \frac{1}{m^2n^2} = \sum_{n=1}^{\infty} \sum_{m=n+1}^{\infty} \frac{1}{m^2n^2} = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{(n+m)^2n^2}$$

Similarly:

$$\zeta(3,1) = \sum_{m=2}^{\infty} \sum_{n=1}^{m-1} \frac{1}{m^3n} = \sum_{n=1}^{\infty} \sum_{m=n+1}^{\infty} \frac{1}{m^3n} = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{(n+m)^3n}$$

Our goal is to prove that

$$\sum_{n,m=1}^{\infty} \frac{1}{n^2m^2} = \sum_{n,m=1}^{\infty} \Big(\frac{4}{(n+m)^3n} + \frac{2}{(n+m)^2n^2}\Big)$$

With finite amount of work we get

$$\frac{4}{(n+m)^3n} + \frac{2}{(n+m)^2n^2} - \frac{1}{n^2m^2} = \cdots = \frac{(m-n)^3}{(n+m)^3n^2m^2}$$

The sum over this is zero due to the antisymmetricity with respect to $n$ and $m$. In other words, the sums over the regions $n>m$ and $n<m$ cancel each other.