Riemann Zeta Function Manipulation

Question

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

Answer 1

(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' applied to $f(x)=x^{-z}$ to prove the existence of a real '$c$' verifying $\,2n-1\le c \le 2n\,$ and such that : $$f'(c)=\frac{f(2n)-f(2n-1)}1$$

Since $f'(c)=-z\,c^{-z-1}\,$ we have : $$f(2n-1)-f(2n)=\frac z{c^{z+1}}$$ getting the upper bound : $$|f(2n-1)-f(2n)|\le \left|\frac z{(2n-1)^{z+1}}\right|$$

and the majoration of our alternate series : $$\left|\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^z}\right|\ \le\ \sum_{n=1}^\infty \;\left|\frac z{(2n-1)^{z+1}}\right|$$

The right part 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.