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
(i) For $\Re(z)> 1$ we have : $$ \begin{align} \zeta(z)&=\sum_{n=1}^\infty \frac 1{n^z}\\ &=\sum_{n=1}^\infty \frac {(-1)^{n+1}}{n^z}+2\sum_{n=1}^\infty \frac 1{(2n)^z}\\ &=\sum_{n=1}^\infty \frac {(-1)^{n+1}}{n^z}+2^{1-z}\,\zeta(z)\\ \end{align} $$ and the result :
$$(1-2^{1-z})\zeta(z)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^z}$$
(ii) For $\Re(z)> 0$ :
$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^z}=\sum_{n=1}^\infty \frac 1{(2n-1)^z} -\frac 1{(2n)^z}$$
You may use the 'mean value theorem' (extended to holomorphic functions) applied to the real and imaginary part of $\,f(x)=x^{-z}\,$ to prove the existence of two real values '$u$' and '$v$' verifying $\,2n-1\le u, v \le 2n\,$ and such that : $$\Re f'(u)=\Re \frac{f(2n)-f(2n-1)}1$$ $$\Im f'(v)=\Im \frac{f(2n)-f(2n-1)}1$$
Since $\,\Re f'(u)=- \Re\left(z\,u^{-z-1}\right)\,$ and $\,\Im f'(v)=- \Im\left(z\,v^{-z-1}\right)\,$ and $\,2n-1\le u, v\;$ we have :
$$|f(2n-1)-f(2n)|^2=\Re^2 (f(2n)-f(2n-1))+\Im^2 (f(2n)-f(2n-1))\\=\Re^2\left(\frac z{u^{z+1}}\right)+\Im^2\left(\frac z{v^{z+1}}\right)\\ \le \left|\frac z{u^{z+1}}\right|^2+\left|\frac z{v^{z+1}}\right|^2\\ \le 2\left|\frac z{(2n-1)^{z+1}}\right|^2$$
getting the upper bound : $$|f(2n-1)-f(2n)|\le \sqrt{2}\left|\frac z{(2n-1)^{z+1}}\right|$$
and the majoration of our alternate series : $$\left|\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^z}\right|\ \le\ \sqrt{2}\sum_{n=1}^\infty \;\left|\frac z{(2n-1)^{z+1}}\right|$$
The series at the right is simply $\ \displaystyle f(x)=\sum_{n=1}^\infty \frac {|z|}{(2n-1)^{x+1}}\ $ with $\ x:=\Re(z)$.
Using the integral test with the observation that $\displaystyle\int_1^\infty \frac {dn}{\,(2n-1)^{x+1}}=\left[\frac {-1}{2x\,(2n-1)^x}\right]_{n=1}^\infty=\frac 1{2x} $ for $x > 0$ we conclude that the alternate series is convergent for $\Re(z)> 0$.
For more about Dirichlet series see this Wikipedia link.