Exploring the Summation of \sum_{n=1}^{ \infty} \frac{1}{n^3}

[jason17349]

I'm interested in the problem:

[tex] \sum_{n=1}^{ \infty} \frac{1}{n^3} [/tex]

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.

 

[dodo]

Sloane sequence A002117 may be of help:
http://www.research.att.com/~njas/sequences/A002117

 

[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!)

 

[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 [tex]pi^3[/tex] is interesting.

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


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

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:

[tex] \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}[/tex]

then:

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

Now I have no idea what [tex]cos(\theta)[/tex] 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 [tex]cos(\theta)\le1[/tex] then:

[tex]\sum_{n=1}^{ \infty} \frac{1}{n^3}\le \frac{\pi^3}{\sqrt{6*90}}[/tex]

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

 

[Gib Z]

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

 

[CRGreathouse]

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