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By an analysis of the matrix of Eulerian numbers(see pg 8) I came across the representation for the alternating Dirichlet series $\eta$: $$ \eta(s) = 2^{s-1} \sum_{c=0}^\infty \left( \sum_{k=0}^c(-1)^k \binom{1-s}{c-k}(1+k)^{-s} \right) \tag 1$$ The H.Hasse/ K.Knopp-form as globally convergent series (see wikipedia) is $$\eta(s) = \sum_{c=0}^\infty \left( { 1\over 2^{c+1} } \sum_{k=0}^c (-1)^k \binom{c}{k}(1+k)^{-s} \right) \tag 2 $$ (Here I removed the leading factor of the $\zeta$-notation in the wikipedia to arrive at the $\eta$-notation)

The difference in the formulae, which made me most curious is that in the binomial-expression, whose upper value is constant in the first formulaand then the same effect in the power-of-2 expression.

I just tried to find a conversion from(1) to (2) but it seems to be more difficult than I hoped. Do I overlook something obvious here? Surely there must be a conversion since the first formula comes from that Eulerian-triangle and this is connected to the sums-of-like powers, but I hope there is an easier one...

Q: "How can the formula (1) be converted into the form (2) ?" or: "how can the equivalence of the two formulae be shown?"


The first formula can be evaluated using the "sumalt"-procedure in Pari/GP which allows to sum some divergent, but alternating series. Here is a bit of code:

myeta(s) = 2^(s-1)*sumalt(c=0,sum(k=0,c,(-1)^k*binomial(1-s,c-k)*(1+k)^(-s)))
myzeta(s)= myeta(s)/(1-2^(1-s))

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I don't understand what your asking, are you asking for someone to show the equivilence of (1) and (2) ? –  Ethan Jan 19 '13 at 2:01
    
@Ethan: well, sometimes finite and also infinite sums involving binomial coefficients cofactored with fractions can be expressed in closed form or in some simpler fashion. That formula (1) is valid is so far only a heuristic (though a very good one). If I would see a nice conversion from (1) to (2) that would be equivalent to a proof of (1) which I could use for my discussion to which I've linked to. –  Gottfried Helms Jan 19 '13 at 2:12
    
@Ethan: so well... ;-). See the edit of my question –  Gottfried Helms Jan 19 '13 at 2:17
    
First id try to break the first sum into even and odd parts , and also get rid of that generalized binomial coeiffient and replace it with a pochhammer symbol or write out the product explictly , and then fiddle with that. –  Ethan Jan 19 '13 at 2:18

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