# Series Name?

1. Sep 17, 2007

### jason17349

I'm interested in the problem:

$$\sum_{n=1}^{ \infty} \frac{1}{n^3}$$

and would like to know more about what attempts have been made at it and any insights into it but I am unable to find much because I don't know the name of this series or if it even has one.

I have learned what little I could about this summation and it's history by searching for the Basel Problem and Reimann Zeta Function but I would really like to find more work related specifically to the above summation. Thanks.

2. Sep 18, 2007

### dodo

3. Sep 18, 2007

### kaisxuans

for this type of questions, if you really want to learn about it, i recommend you start with the basics and not jump into an advanced question just yet. try searching the web or something. it could prove useful.
(viait my blog!)

4. Sep 18, 2007

### jason17349

Kaisxuans - I had no plan to jump into the dificult Zeta Function but it's one of the few things related to my problem that I can search for. Other than that I don't know what to look for.

Dodo - Thanks for the link. This first part about wether Zeta(3) is a rational multiple of $$pi^3$$ is interesting.

I was tinkering with this problem the other day when I tried representing it as a multiple of two vectors.

$$\sum_{n=1}^{ \infty} \frac{1}{n^3} =$$[1 1/2 1/3 ....]* [1 1/2^2 1/3^2 ....]$$^T$$

which can be written as the magnitude of the first vector times the magnitude of the second vector times the cos of the angle between them.

which works out to:

$$\sum_{n=1}^{ \infty} \frac{1}{n^3}=\sqrt{\sum_{n=1}^{ \infty} \frac{1}{n^2} }*\sqrt{\sum_{n=1}^{ \infty} \frac{1}{n^4}}*\cos{\theta}$$

then:

$$\sum_{n=1}^{ \infty} \frac{1}{n^3}=\frac{\pi^3}{\sqrt{6*90}}*cos(\theta)$$

Now I have no idea what $$cos(\theta)$$ is and can't prove if it's rational but I was wondering if this method had ever been used before and if anybody had gotten any farther with it. An interesting consequence (at least I thought it was interesting) is that since $$cos(\theta)\le1$$ then:

$$\sum_{n=1}^{ \infty} \frac{1}{n^3}\le \frac{\pi^3}{\sqrt{6*90}}$$

but this is trivial because it becomes obvious after just a few terms that the summation is well below $$\frac{\pi^3}{\sqrt{6*90}}$$.

Last edited: Sep 18, 2007
5. Sep 22, 2007

### Gib Z

Mclosed form expression has yet been found for the series, although many integral and series representations are available.

6. Sep 22, 2007

### CRGreathouse

Calculating zeta(3), dividing out $\pi^3$, and calculating a continued fraction expansion shows that it it's a rational multiple the denominator is huge.