Riemann function variant: odd numbers only? I apologize for being not able to provide much context.
Is there a special zeta function defined as:
$$f(s)=\sum^\infty_{n=1}\frac1{(2n-1)^s}$$
?
Moreover, if I know the value of $f(k)$($k$ is an integer), can we thus find the value of $\zeta(k)$?
I know very little about this topic, and recently am facing difficulties, so any help is highly appreciated.
Thanks in advance.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\left.\rule{0pt}{5mm}\mrm{f}\pars{s}\right\vert_{\ \Re\pars{s}\ >\ 1} & \equiv \sum_{n = 1}^{\infty}{1 \over \pars{2n - 1}^{s}} =
\sum_{n = 1}^{\infty}\bracks{{1 \over n^{s}} - {1 \over \pars{2n}^{s}}} =
\pars{1 - 2^{-s}}\sum_{n = 1}^{\infty}{1 \over n^{s}}
\\[5mm] & =
\bbx{\pars{1 - 2^{-s}}\zeta\pars{s}}
\end{align}
