Can anyone compute this series?

[Johnny5]

I'm not sure what different methods there are out there to compute the following series:

 

[math] \sum_{n=0}^{n=\infty} \frac{(-1)^n}{(2n+1)^3} [/math]

[matt grime]

It looks to me like the coefficients of some fourier series.

[lurflurf]

The sum is pi^3/32~.968946146259

It is the Dirichlet Beta Function evaluated at 3.

http://mathworld.wolfram.com/DirichletBetaFunction.html

[Johnny5]

Ok pi^3/32, is there an easy way to show this? How exactly does pi enter the answer?

[matt grime]

Why shouldn't pi enter into it? If it's anything to do with integrals of any reasonably nice function then pi is bound to play a role. Learn about, ooh, fourier series to see why.

[Johnny5]

Learn about, ooh, fourier series to see why.

 

Ok, I've done Fourier series before, and now that you mention it, I recall that you can get a whole bunch of series in terms of pi. It was in my PDE book, but i dont know where that is at the moment. But i know what you're talking about.

 

But, off the top of my head, I do not remember the entire comprehensive analysis given in that PDE book.

 

I remember off the top of my head some of what has to be done.

 

There are a series of integrals that you have to understand.

 

sin(mx)cos(nx)dx

sin(mx)sin(nx)dx

cos(mx)cos(nx)dx

 

And as i recall, the answers are either zero or one. And I solved them using integration by parts.

 

Then lets see.

 

The idea behind fourier series goes like this.

 

Suppose that you have some function f(x).

 

Suppose next that it has a Fourier series expansion on some interval L.

 

So we have something like this going on:

 

-L/2 < x < L/2

 

 

Possible Fourier sine series:

 

[math] f(x) = \sum_{n=0}^{n=\infty} c_n sin (nx) [/math]

 

Possible Fourier cosine series:

 

[math] f(x) = \sum_{n=0}^{n=\infty} c_n cos (nx) [/math]

 

Now, you have to work out the formulas for the Fourier coefficients, which is what takes the most time.

 

All that was without googling.

[lurflurf]

Easy way is subjective. I did find a function whose fourier series in cos that can be used to find the sum.

f(x)=pi^3/32-(pi^3/16)x^2 0<x<.5

f(x)=pi^3/64-(pi^3/16)(x-.5)^2 .5<x<1

f(x)=(pi^3/16)(x-1)^2 1<x<1.5

f(x)=pi^3/64+(pi^3/16)(x-1.5)^2 1.5<x<2

find cos series

the sum is f(0)=pi^3/32

 

A few easy things to try when finding sums

1. recognize it in terms of a known function

my first thought when I saw it was sum=im(trilog(i))

but that was no easier than the sum, the integrals just gave back polylogs giving no progress.

Next I thought of it as one of the functions related to zeta in fact Dirichlet Beta this gave the sum easily.

The above function also gives the sum.

2. constuct a function have residues equal to terms of the sum so that the sum can be calculated by contour integration.

That did not seem to good for this.

3. add a parameter then relate the sum to a known quantity and solve for it

as in 1 I imbeded the sum in trilog(x) a 3rd order linear ode was apparent but knowing the solution was trilog(x) was not helpful.

 

Any of those three approches make it clear why pi occurs in so many sums. I'm sure there are some good books on doing sums out there.

[lurflurf]

The above function is an error.

f(x)=(pi^3/8)(x-x^2) expanded in sine over0<x<1 gives a series that can be used

sum=f(1/2)=pi^3/32

This function can be deduced by thinking about the sine expansion of 1 on the same interval and integrating twice with respect to x.

 

Here it is with trilog(x) using a helpful identity from

http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/PolyLog/17/02/03/01/

(trilog(1/z)-trilog(z))2=log^3(-z)/12+(pi^2)log(-z)/12

-i sum=(tilog(-i)-trilog(i))/2

-i sum=(tilog(1/i)-trilog(i))/2

-i sum=log^3(-i)/12+(pi^2)log(-i)/12

-i sum=(-pi i/2)^3/12+(pi^2)(-pi i/2)/12

-i sum=pi^3 i/96-pi^3 i /24

-i sum=-i pi^3/32

sum=pi^3/32