It is well known the common Riemann Zeta series $\sum 1/n^s$, converges for $\sigma>1$ where $s=\sigma+it$.
The following series extends the domain to be convergent over $\sigma>0$.
$$\boxed{\zeta(s)=\frac{1}{1-2^{1-s}}\sum\frac{(-1)^{n+1}}{n^{s}}}$$
Aim: I want to prove this has only 1 pole at $s=1$.
My Argument
The alternating zeta function, also known as the eta $\eta(s)$ function is as follows:
$$\eta(s)=\sum\frac{(-1)^{n+1}}{n^{s}}=1-\frac{1}{2^{s}}+\frac{1}{3^{s}}-\frac{1}{4^{s}}+\frac{1}{5^{s}}-\ldots$$
Dirichlet theory (eg Apostol's IANT) allows us to observe that because $\sum (-1)^{n+1}$ is bounded, the series converges for $\sigma>0$ where $s=\sigma+it$.
That means $\eta(s)$ has no poles in the domain $\sigma>0$ and therefore the above $\zeta(s)$ must get any poles from the factor $1/(1-2^{1-s})$.
That factor diverges when $1-s=0$, which means $s=1$. Therefore the above $\zeta(s)$ has only one pole at $s=1$.
Question: Is this argument correct and sufficient?