Convergence of alternting series of complex numbers in Riemann Zeta [duplicate]

Solution 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.