Given $\displaystyle \displaystyle \text{Re}(s)>0 $ and $\displaystyle \displaystyle x>0 $, show $\displaystyle \displaystyle \zeta(s)=\frac1{1-2^{1-s}}\sum_{n=0}^\infty\sum_{k=0}^n\binom{n}{k}\frac{ x^{n-k}}{(1+x)^{n+1}}\frac{(-1)^k}{(k+1)^s} $.

Printable View

- Aug 17th 2010, 09:41 PMchiph588@Riemann Zeta Function
Given $\displaystyle \displaystyle \text{Re}(s)>0 $ and $\displaystyle \displaystyle x>0 $, show $\displaystyle \displaystyle \zeta(s)=\frac1{1-2^{1-s}}\sum_{n=0}^\infty\sum_{k=0}^n\binom{n}{k}\frac{ x^{n-k}}{(1+x)^{n+1}}\frac{(-1)^k}{(k+1)^s} $.

- Aug 18th 2010, 11:30 AMVlasev
Chiph, I believe that the series should be:

$\displaystyle \displaystyle \frac{1}{1-2^{1-s}}\sum _{n=0}^{\infty } \sum _{k=0}^n \frac{(-1)^k

\binom{n}{k} x^{n-k}}{(k+1)^s (x+1)^{n+1}}}$

Here is my solution:

__Spoiler__: - Aug 21st 2010, 04:07 AMVlasev
Where did you find this identity?

- Aug 22nd 2010, 08:42 PMchiph588@
- Aug 22nd 2010, 09:53 PMVlasev
There is one online but it's just a special case of this one. It's for x = 1. They do reference the paper where it was derived and I'm sure they did it for arbitrary x. Maybe that's what you saw. Thanks though.

I think the series could be derived with a straight application of Euler summation: Euler summation - Wikipedia, the free encyclopedia