The Riemann zeta function is defined on the $Re z> 1$ by $\zeta(z)=\sum_{n=1}^\infty \frac{1}{n^z}$
(i) show that for $Re z> 1$, we have $(1-2^{1-z})\zeta(z)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^z}$%
(ii) show that $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^z}$ is an analytic function on $Re z> 0$
Thoughts thus far:
(i) Since $2^z=2^{Rez+iImz}=2^{Rez}2^{iImz}=2^{Rez}[\cos(Imz\ln2)+i\sin(Imz\ln2)]$ we obtain $(1-2^{1-z})\zeta(z)=\sum_{n=1}^\infty \frac{1}{n^z}-\sum_{n=1}^\infty \frac{2^{1-z}}{n^z}=\sum_{n=1}^\infty \frac{1}{n^z}-\sum_{n=1}^\infty \frac{2}{2^{Rez}[\cos(Imz\ln2)+i\sin(Imz\ln2)]n^z}=$ (by multiplying by the conjugate) $\sum_{n=1}^\infty \frac{1}{n^z}-\sum_{n=1}^\infty \frac{2(\cos(Imz\ln2)-i\sin(Imz\ln2))}{2^{Rez}[\cos^2(Imz\ln2)+\sin^2(Imz\ln2)]n^z}=$ (since $\sin^2\theta+\cos^2\theta=1$) $\sum_{n=1}^\infty \frac{1}{n^z}-\sum_{n=1}^\infty \frac{2(\cos(Imz\ln2)-i\sin(Imz\ln2))}{2^{Rez}n^z}=\sum_{n=1}^\infty \frac{2^{Rez}-2(\cos(Imz\ln2)-i\sin(Imz\ln2))}{2^{Rez}n^z}$, at which point I get stuck. I am uncertain whether unraveling $2^z$ in the manner that I did was fruitful or perhaps, as usual, I am missing something berry basic.
(ii) Since we want to show analyticity, we may show that the power series converges. Using the logic as above, we know that $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^z}=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^{Rez}[\cos(Imz\ln n)+i\sin(Imz\ln n)]}=\sum_{n=1}^\infty \frac{(-1)^{n+1}[\cos(Imz\ln n)-i\sin(Imz\ln n)]}{n^{Rez}}$ Since $\cos(Imz\ln n)-i\sin(Imz\ln n)$ represents oscillations around the unit circle we may clearly see that $\lim_{n \to \infty}|\frac{(-1)^{n+1}\cos(Imz\ln n)-i\sin(Imz\ln n)}{n^{Rez}}|=\lim_{n \to \infty}|\frac{1}{n^{Rez}}|=0\iff Rez>0$ and hence the power series converges if and only if Rez>0. Does this appear to be a valid proof for analyticity?
Thanks in advance for any help that you may provide