Can that two double series representations of the $\eta$/$\zeta$ function be converted into each other? 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))


 A: For $|z|<1$ and any $s$
$$-Li_s(-z) (1+z)^{1-s}= \sum_k z^k (-1)^{k+1}k^{-s}\sum_m z^m {1-s\choose m}=\sum_c z^c \sum_{k\le c} (-1)^{k+1}k^{-s}{1-s\choose c-k}$$
Interpret $-Li_s(-e^{2\pi  it}) (1+e^{2\pi it})^{1-s}$ as $\lim_{r\to 1^-}-Li_s(-r e^{2\pi  it}) (1+r e^{2\pi it})^{1-s}$.

With enough partial summations we have that $Li_s(z)$ is continuous for $|z|\le 1,z\ne 1$ and for $\Re(s)\le 1,t\not \in \Bbb{Z}$, the value on the boundary $Li_s(e^{2i\pi t})$ is the  analytic continuation of $Li_s(e^{2i\pi t}),\Re(s) > 1$.

Also $-Li_s(-e^{2\pi  it}) (1+e^{2\pi it})^{1-s}\in L^1(\Bbb{R/Z})$, thus $\sum_c e^{2i\pi t c} \sum_{k\le c} (-1)^{k+1}k^{-s}{1-s\choose c-k}$ is its Fourier series.

Your claim is that the Fourier series converges at $t=0$,

Which is true, because $-Li_s(-e^{2\pi  it}) (1+e^{2\pi it})^{1-s}$ is $C^1$ at $t=0$.

Whence for all $s\in \Bbb{C}$ $$\sum_{c=0}^\infty \sum_{k\le c} (-1)^{k+1}k^{-s}{1-s\choose c-k}=2^{1-s}\eta(s)$$

I think it is quite different to the spirit of the proof of (2)
