TITLE: Alternating series test for complex series
QUESTION [8 upvotes]: 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?

REPLY [6 votes]: 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!