Prove that $(1 - 2^{1-z})\zeta(z) = 1^{-z} - 2^{-z} + 3^{-z} - \cdots$

Question

Prove that $$(1 - 2^{1-z})\zeta(z) = 1^{-z} - 2^{-z} + 3^{-z} - \cdots$$ and that the latter series represents an analytic function for Re $z > 0.$

I think I can use the following theorem for the second part, but I'm not seeing how I'm supposed to prove the equality.

If a series with analytic terms, $$f(z) = f_1(z) + f_2(z) + \cdots+ f_n(z) + \cdots,$$ converges uniformly on every compact subset of a region $\Omega,$ then the sum $f(z)$ is analytic in $\Omega$, and the series can be differentiated term by term.

Answer 1

No need to use such complicated method.$$(1-2^{1-z})\zeta(z){\\=\sum_{n=1}^{\infty}\dfrac{1}{n^s}-2\sum_{n=1}^{\infty}\dfrac{1}{(2n)^s}\\=\sum_{\text{n is odd}}\dfrac{1}{n^s}+\sum_{\text{n is even}}\dfrac{1}{n^s}-2\sum_{\text{n is even}}\dfrac{1}{n^s}\\=\sum_{\text{n is odd}}\dfrac{1}{n^s}-\sum_{\text{n is even}}\dfrac{1}{n^s}\\=\dfrac{1}{1^s}-\dfrac{1}{2^s}+\dfrac{1}{3^s}-\dfrac{1}{4^s}+\cdots}$$

Answer 2

\begin{eqnarray*} \zeta(z)= 1+\frac{1}{2^z} +\frac{1}{3^z}+ \cdots \\ 2^{1-z} \zeta(z)= 2\left(\frac{1}{2^z} +\frac{1}{4^z}+ \frac{1}{6^z} +\cdots \right) \\ \end{eqnarray*} Subtract these & the result follows.