Alternating series test for complex series I want to show that we can continue Riemann's zeta function to Re$(s)>0$, $s\neq 1$ by the following formula
\begin{align}
(1-2^{1-s})\zeta(s)&=\left(1-2\frac{1}{2^s}\right)\left(\frac1{1^s}+\frac1{2^s}+\ldots \right) \\
&=\frac1{1^s}+\frac1{2^s}+\ldots -2\left(\frac1{2^s}+\frac1{4^s}+\ldots \right)\\
&=\frac1{1^s}-\frac1{2^s}+\frac1{3^s}-\frac1{4^s}+\ldots  \\
&=\sum_{n=1}^\infty (-1)^{n-1}\frac1{n^s}.
\end{align}
In order to do that, I need to show that the series converges for Re$(s)>0$, except $s=\frac{2k\pi i}{\ln 2}+1$, $k\in \mathbb{Z}$, which are removable singularities. Any ideas on how to do that?
 A: 
Here, I thought it might be instructive to present an approach that uses a generalization of Dirichlet's Test and that has wider applicability.  To that end we proceed.

Let $s\in \mathbb{C}$.  The Dirichlet Eta function, $\eta(s)$, as represented by the series 
$$\eta(s)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} \tag 1$$
is easily seen to converge for $\text{Re}(s)>1$. 
If $s$ is purely real, then Dirichlet's test guarantees that the series representation converges for $\text{Re}(s)=s>0$.  If $s$ is not purely real, then Dirichlet's test is inapplicable since the term $\frac{1}{n^s}$ is not real and monotonically decreasing.   
It is not obvious, therefore, that the series in $(1)$ converges when $\text{Re}(s)>0$ for general complex values of $s$.  There is a generalization of Dirichlet's test See Here to which we can appeal and show the convergence of $(1)$ whenever $\text{Re}(s)>0$.  

The Generalized Dirichlet Test states that if $a_n$ and $b_n$ are, in general, complex sequences, then the sequence of their product, $\sum_{n=1}^\infty a_nb_n$, converges under the following three conditions:
  
  
*
  
*There exists a number $M$, independent of $N$, such that the partial sums of $b_n$ satisfy 
  
  
  $$\left|\sum_{n=1}^N b_n\right|\le M$$

  
*The terms $a_n$ tend to zero as $n\to \infty$
  
*The sequence $a_n$ is of bounded variation with $\sum_{n=1}^\infty |a_{n+1}-a_n| \le L <\infty$

Let $a_n=\frac{1}{n^s}$ and $b_n=(-1)^{n-1}$, $\text{Re}(s)>0$.  Clearly conditions $1$ and $2$ are satisfied.  To show that $3$ is satisfied, we note that for fixed $s$ with $\text{Re}(s)>0$
$$\begin{align}
\sum_{n=1}^\infty \left|\frac{1}{(n+1)^s}-\frac{1}{n^s}\right|& \le \int_1^\infty \left|\frac{d}{dt}\left(\frac{1}{t^s}\right)\right|\,dt\\\\
&=\int_1^\infty \left|\frac{-s}{t^{s+1}}\right|\,dt\\\\
&=|s|\int_1^\infty \frac{1}{t^{1+\text{Re}(s)}}\,dt\\\\
&=\frac{|s|}{\text{Re}(s)}
\end{align}$$
where we recognized that $\sum_{n=1}^\infty \left|\frac{1}{(n+1)^s}-\frac{1}{n^s}\right|$ represented the sum of the lengths of secant lines of the parametric curve $\frac{1}{t^s}$, $t\in [1,\infty)$.
Therefore, by the Generalized Dirichlet Test, the series 
$$\eta(s) =\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}$$
converges when $\text{Re}(s)>0$.  And we are done!
A: *

*Write the Taylor expansion of $(1+x)^a$ around $x= 0$, and $$n^{-s} - (n+1)^{-s} = n^{-s}(1- \left(\frac{n+1}{n}\right)^{-s}) = n^{-s}(1- \left(1+\frac{1}{n}\right)^{-s}) = \mathcal{O}(n^{-s-1})$$
which is summable for $Re(s) > 0$, hence by grouping the terms by two :
$$\eta(s)  = (1-2^{1-s})\zeta(s) = \sum_{n=1}^\infty (2n-1)^{-s} - (2n)^{-s}$$ is absolutely convergent for any $Re(s) > 0$. ($\eta(s)$ is called the Dirichlet $\eta$ function )

*Other way, integrate by part (with $\delta(x)$ the Dirac delta distribution)$$\sum_{n=1}^\infty a_n n^{-s} = \int_{1-\epsilon}^\infty \left(\sum_{n=1}^\infty a_n \delta(x-n)\right) x^{-s} dx = s \int_1^\infty \left(\sum_{n \le x} a_n\right) x^{-s-1} dx$$ here $a_n = (-1)^{n+1}$ hence $\sum_{n \le x} a_n = 1$ or $0$ and $$\eta(s) = s \int_1^\infty \left(\sum_{n \le x} (-1)^{n+1}\right) x^{-s-1} dx$$ converges for $Re(s) > 0$ (it's called the Perron's formula )

*Last way, use that $n^{-s} \Gamma(s) = n^{-s} \int_0^\infty x^{s-1} e^{-x} dx = \int_0^\infty y^{s-1} e^{-ny} dy$ (change of variable $y = nx$, and $\Gamma(s)$ the Gamma function) hence :
$$\Gamma(s) \sum_{n=1}^\infty a_n n^{-s} = \int_0^\infty y^{s-1} \left(\sum_{n=1}^\infty a_n e^{-ny} \right) dy$$
(by using the absolute/dominated convergence theorem for exchanging $\sum$ and $\int$)
here $\sum_{n=1}^\infty (-1)^{n+1} e^{-ny} = \frac{1}{e^y+1}$ hence
$$\eta(s) \Gamma(s) = \int_0^\infty \frac{y^{s-1}}{e^y+1} dy$$
which has no singularity for $Re(s)  > 0$.

*Last last way : $$\eta(\epsilon) = \sum_{n=1}^\infty (-1)^{n+1} n^{-\epsilon}$$
is a convergent alternated series for $\epsilon > 0$, hence by the abscissa of convergence theorem for Dirichlet series, we get that $\sum_{n=1}^\infty (-1)^{n+1} n^{-s}$ converges for any $Re(s) > \epsilon$, and since $\epsilon$ is arbitrary small, for any $Re(s) > 0$.
(if someone knows another way...)
