Challenge 1: Multiple Zeta Values

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?

 

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.

 

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

 

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.

 

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.

 

Well, that's an intriguing remark... OK, I'll try to think about this a bit more tonight.

 

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.

 

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.

 

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

 

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.

 

We follow Office_Shredder's latest advice:

[tex]
\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}
[/tex]

Similarly:

[tex]
\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}
[/tex]

Our goal is to prove that

[tex]
\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)
[/tex]

With finite amount of work we get

[tex]
\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}
[/tex]

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