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))