Analytic continuation of Riemann zeta-function I showed that 
$$\zeta(s)=\frac1{1-2^{1-s}}\sum_{n=1}^\infty (-1)^{n-1}\frac1{n^s}.$$
Right hand side should be analytic when Re$(s)>0$ and $s\neq 1$, but there are problems at points $s=\frac{2k\pi i}{\ln 2}+1$, $k\in \mathbb{Z}$. How can we get rid of those singularities? Why does $s=1$ remain?
 A: yes there is a problem at $s = 1+\frac{2 i k \pi}{\ln 2}, k \in \mathbb{Z}$. a common solution for solving it :
at first for $Re(s) > 1$, then by analytic continuation for $Re(s) > 0$  :
$$\zeta(s) = \frac{1}{1-3^{1-s}}\sum_{n=0}^\infty (3n+1)^{-s} +(3n+2)^{-s}- 2 (3n+3)^{-s} = \frac{\eta_3(s)}{1-3^{1-s}}$$
where (exactly as $\eta(s) = \sum_{n=1}^\infty (-1)^{n+1} n^{-s}$) the function $\eta_3(s)$ is analytic on $Re(s) > 0$.
hence with this formulation, $\zeta(s)$ is analytic on $Re(s) > 0$ except at $s = 1+\frac{2 i k \pi}{\ln 3}, k \in \mathbb{Z}$, and since $2$ and $3$ are relatively primes, the intersection between $\{ 1+\frac{2 i k \pi}{\ln 2} \ \mid \ k \in \mathbb{Z}\} $ and $\{1+\frac{2 i k \pi}{\ln 3} \ \mid \ k \in \mathbb{Z}\}$ reduces to $\{1\}$, and hence $\zeta(s)$ is analytic on $Re(s) > 0, s \ne 1$.
this is also the way for proving that $\eta(1+\frac{2 i k \pi}{\ln 2}) = 0$ for every $ k \in \mathbb{Z}\setminus \{0\}$.
(this is explained on https://fr.wikipedia.org/wiki/Fonction_zêta_de_Riemann, or you can look at https://en.wikipedia.org/wiki/Dirichlet_eta_function)
from the fact that $1-2^{1-s}$ has a zero of order $1$ at $s=1$, we also get that $(s-1)\zeta(s)$ is analytic at $s=1$, and since $$\lim_{\sigma \to 1^+} \zeta(\sigma) = \lim_{\sigma \to 1^+} \sum_n n^{-\sigma} = + \infty$$
we find that $\zeta(s)$ is not analytic at $s=1$, hence it has a pole of order $1$ there.
A: Consider the series $$F(s)=\frac{1}{1^s}+\frac{1}{2^s}-\frac{2}{3^s}+\frac{1}{4^s}+\frac{1}{5^s}-\frac{2}{6^s}+\cdots$$
First, convergence:
$$
\begin{eqnarray}
F(s)&\leq&\frac{1}{1^s}+\frac{1}{1^s}-\frac{2}{3^s}+\frac{1}{4^s}+\frac{1}{4^s}-\frac{2}{6^s}+\cdots \\
&=&\frac{2}{1^s}-\frac{2}{3^s}+\frac{2}{4^s}-\frac{2}{6^s}+\frac{2}{7^s}\cdots
\end{eqnarray}
$$
This converges for $\mathfrak{R}(s)\geq0$ for the same reasons the series in your formula does.
One quickly checks that $\zeta(s)-F(s)=\frac{3}{3^s}\zeta(s)$, and so $$\zeta(s)=\frac{1}{1-3^{1-s}}F(s)$$
Thus we have an analytic continuation of $\zeta$ with poles at $s=1+k\frac{2\pi i}{\ln(3)}$.  The only common pole shared with your formula is thus at $s=1$.  Together, the two formulas show that $\zeta$ is analytically continuable to $s>0$, except at $s=1$
At $s=1$, no such trick will work.  To see that $\zeta$ has an actual pole there, it suffices to show that the residue of $\zeta$ at $s=1$ is non-zero.  Using the formula you quote:
$$
\begin{eqnarray}
\mathrm{Res}(\zeta,1) &=&\lim_{s\rightarrow 1}(s-1)\zeta(s) \\
&=&\lim_{s\rightarrow 1}\frac{(s-1)}{1-2^{1-s}}\sum_{n=1}^\infty(-1)^n\frac{1}{n^s} \\
&=&\lim_{s\rightarrow1}\frac{(s-1)}{1-2^{1-s}}\ln(2) & \textrm{(Taylor series of }\ln(1+s))\\
&=&\lim_{s\rightarrow1}\frac{1}{(\ln{2})2^{1-s}}\ln(2) & \textrm{(L`Hospital's rule)} \\
&=& 1
\end{eqnarray}
$$
