Properties of $\underset{k\geq1}{\sum}\frac{1}{\left(2k-1\right)^{s}}$

Question

Is this function $$\underset{k\geq1}{\sum}\frac{1}{\left(2k-1\right)^{s}},\,Re(s)>1$$ well known? In particular I'm interessed about analytic continuation and its zeros and poles. Have this function the same properties of the Riemann Zeta function? Thank you.

Answer 1

Formally,$$\zeta\left(s\right)=\sum_{k=1}^{+\infty}\frac{1}{k^{s}}=\sum_{k=1}^{+\infty}\frac{1}{\left(2k\right)^{s}}+\sum_{k=1}^{+\infty}\frac{1}{\left(2k-1\right)^{s}}=\frac{1}{2^{s}}\sum_{k=1}^{+\infty}\frac{1}{k^{s}}+\sum_{k=1}^{+\infty}\frac{1}{\left(2k-1\right)^{s}}$$ whence$$\sum_{k=1}^{+\infty}\frac{1}{\left(2k-1\right)^{s}}=\zeta\left(s\right)\left(1-\frac{1}{2^{s}}\right).$$

Answer 2

Logger the function is well--known: it is the Dirichlet L-function modulo $2$. It is not primitive since $\phi(2)=1$ so it must be induced by the classical zeta function.