How can the numerical value of the infinite Zeta Function sum be found?

[seanhbailey]

Homework Statement

Find the numerical value of [tex]\sum_{k=0}^{\infty} (\zeta(-k))[/tex]

Homework Equations

The Attempt at a Solution

I have no idea how to get a numerical value for this sum.

 

[seanhbailey]

The sum [tex]\sum_k=0^\infty (\zeta(1-2k))[/tex] is equal to [tex]\sum_k=0^\infty (-B_{2k}\2k)[/tex]. I hope this helps. Thanks

 

[seanhbailey]

I forgot to mention that B represents the Bernoulli numbers.

 

[Petek]

I think that

[tex]\zeta(1 - 2k)[/tex] = [tex]\frac{(-1)^{2k-1} B_{2k}}{2k}[/tex].

At least, that's what's in one of my books. Also, is that first sum correct? It looks like you're supposed to use the identity

[tex]\frac{x}{e^x - 1} = \sum_{n=0}^\infty \frac{B_nx^n}{n!}[/tex]

but you need another k in the sum's denominator.

Petek

 

[seanhbailey]

Thank you for helping. Sorry about the first sum; I typed it in wrong.

 

[seanhbailey]

Because of the 1/k in the denominator, does this imply that [tex]\sum_{k=0}^{\infty} (\zeta(-k))[/tex] has no sum?

 

[seanhbailey]

I found that zeta(-k) is equal to B(n)/(((-1)^(n+1))*n), where B(n) is the Bernoulli numbers, implying that [tex]\sum_{k=0}^{\infty} (\zeta(-k))[/tex] is equal to ln(2)*[tex]\sum_{k=0}^{\infty} (B(n))[/tex]. Sorry about the formating.

 

[Petek]

To make sure that we're solving the same problem, please post the sum that you're trying to evaluate (since one of your posts stated that the sum in your original post was inaccurate). Thanks!

Petek

 

[seanhbailey]

The original sum I was trying to evaluate was [tex]\sum_{k=0}^{\infty} (\zeta(-k))[/tex].

 

[seanhbailey]

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