Other series representations for zeta and eta functions

Question

Given that $ \left(\forall s>1\right),\ \zeta\left(s\right)=\sum\limits_{n=1}^{+\infty}{\frac{1}{n^{s}}} $, and that $ \left(\forall s>0\right),\ \eta\left(s\right)=\sum\limits_{n=1}^{+\infty}{\frac{\left(-1\right)^{n-1}}{n^{s}}} $, prove the following formulas :

$$ \left(\forall s>0\right),\ \eta\left(s\right)=\sum_{n=0}^{+\infty}{\frac{1}{2^{n+1}}\sum_{k=0}^{n}{\left(-1\right)^{k}\binom{n}{k}\left(k+1\right)^{-s}}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(1\right) $$

$$ \left(\forall s>1\right),\ \zeta\left(s\right)=\frac{1}{s-1}\sum_{n=0}^{+\infty}{\frac{1}{n+1}\sum_{k=0}^{n}{\left(-1\right)^{k}\binom{n}{k}\left(k+1\right)^{1-s}}} \ \ \ \left(2\right) $$

These formulas appear to be algorithms for accelerating the convergence of these series.

Answer 1

Use Euler's transform (ET)

$$\sum_{n\geq0}(-1)^na_n=\sum_{n\geq0}\frac{(-1)^n}{2^{n+1}}\sum_{k=0}^n(-1)^k\binom nka_{n-k}$$

For the Dirichlet Eta Function we obtain

\begin{align*} \eta(s)=\sum_{n\geq1}\frac{(-1)^{n+1}}{n^s}&=\sum_{n\geq0}(-1)^na_n&&;~a_n=(n+1)^{-s}\\ &=\sum_{n\geq0}\frac{(-1)^n}{2^{n+1}}\sum_{k=0}^n(-1)^k\binom nka_{n-k}&&;~\text{ET}\\ &=\sum_{n\geq0}\frac{(-1)^n}{2^{n+1}}\sum_{k=0}^n(-1)^k\binom nk(n-k+1)^{-s}&&;~n-k\mapsto k\\ &=\sum_{n\geq0}\frac{(-1)^n}{2^{n+1}}\sum_{k=0}^n(-1)^{n-k}\binom n{n-k}(k+1)^{-s}\\ &=\sum_{n\geq0}\frac1{2^{n+1}}\sum_{k=0}^n(-1)^k\binom nk(k+1)^{-s} \end{align*}

The accelerated series for the Riemann Zeta Function is listed here and a proof may be found here (Theorem $10.$ in particular). As far as I can tell the idea is similiar to the one given for $\eta(s)$.